Solving Trig Equations Using Identities

Solution: \(x=\dfrac{\pi}{3},\dfrac{5\pi}{3}\)

Solution Method: This is a case where we’ll use multiple trig identities to solve a trigonometric equation. In this case, cofunction and even/odd identities. Cofunction identities are a special case of the sum/difference identites. Even/Odd identities derive from cosine being an even function and sine and tangent being odd functions. You can easily verify that these functions and even and odd respectively by looking at their graphs. Cosine has y-axis symmetry, so it’s even. Sine and tangent have origin symmetry, and are odd.

Comparing my expression to the identities, I notice that \(\sin\left( -\frac{\pi}{2}+x \right)\) is very close to the cofunction identity \(\cos{x}= \sin\left( \frac{\pi}{2}-x \right)\). I just need to distribute out a negative to make them match. \[\begin{aligned} -\frac{1}{2}&=\sin\left( -\frac{\pi}{2}+x \right) \\ -\frac{1}{2}&=\sin\left( -(\frac{\pi}{2}-x) \right) \end{aligned}\] Then by the odd identity of sine, that negative can come out of the argument. \[\begin{aligned} -\frac{1}{2}&=-\sin\left( \frac{\pi}{2}-x\right) \end{aligned}\] So now I can apply the cofunction identity, \[\begin{aligned} -\frac{1}{2} &= -\cos{x} \\ \frac{1}{2} &= \cos{x} \end{aligned}\] So now we’re looking for values of x in the interval \([0,2\pi]\) that satisfy this. That interval is just the unit circle so I need to remember what special angles produce cosine of one-half on the unit circle. First, this is a positive half so I can only have angle values in Q1 and Q4. Cosine corresponds to the x-values on the unit circle and I know that my x-coordinate is one-half for reference angle values of \(\frac{\pi}{3}\). In Q1, that is just \(\frac{\pi}{3}\), in Q4 it is \(\frac{5\pi}{3}\). So the values of x in \([0, 2\pi]\) satisfying the equation are \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\).

Solution: \(x = \dfrac{\pi}{3} + \dfrac{2\pi n}{3} , n \in\mathbb{Z}\)

Solution Method: This equation has both sine and cosine terms and differing powers, which makes it complicated to solve. It is easier to solve an equation of one trig value. This is where trig identities come in. Specifically, the Pythagorean identity \(\sin^2{x}+\cos^2{x}=1\), relates sine and cosine. I have a \(\sin^2{x}\) in my equation, so I can solve the identity for \(\sin^2{x}\) in terms of \(\cos{x}\), and substitute in to the equation. \[\begin{aligned} \sin^2{x}+\cos^2{x}=1 \\ \sin^2{x}=1-\cos^2{x} \end{aligned}\] \[\begin{aligned} 2\sin^2{x}&=1+\cos{x} \\ 2(1-\cos^2{x})&=1+\cos{x} \\ 2-2\cos^2{x} &= 1+\cos{x} \end{aligned}\] Now I have a quadratic of cosine, so I’ll move all terms onto one side so my quadratic is equal to zero. \[\begin{aligned} -2\cos^2{x}-\cos{x}+1 &=0 \\ 2\cos^2{x}+\cos{x}-1 &=0 \end{aligned}\] Now I’ll try to factor the quadratic of cosine like I would the equation \(2x^2+x-1=0\). So I need two numbers that multiply to be -2 and sum to 1. So they must be 2 and -1.

Since the leading coefficient of the quadratic is not one, we will split \(\cos{x}\) into \(2\cos{x}\) and \(-\cos{x}\) and factor by grouping. \[\begin{aligned} 2\cos^2{x}+\cos{x}-1 &=0 \\ 2\cos^2{x}+2\cos{x}-\cos{x}-1 &= 0 \\ 2\cos{x}(\cos{x}+1)-(\cos{x}+1) &= 0 \\ (2\cos{x}-1)(\cos{x}+1) &= 0 \end{aligned}\] So now I can set each factor to 0 separately and solve for x. \[\begin{aligned} 2\cos{x}-1&=0 \\ 2\cos{x}&=1 \\ \cos{x} &= \frac{1}{2} \end{aligned}\] The reference angle that produces cosine equal to \(\frac{1}{2}\) is \(\frac{\pi}{3}\). The one-half is positive. Cosine is positive in (ASTC) Q1 and Q4. So the values satisfying \(\cos{x}=\frac{1}{2}\) on \([0,2\pi]\) are \(\frac{\pi}{3}\) in Q1 and \(\frac{5\pi}{3}\) in Q4. But I was asked to find all the values of x satisfying the equation. Since trig functions are periodic, they repeat themselves every rotation, every \(2\pi\). So every full rotation, \(2\pi\), from \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\) is another value satisfying the equation. So we can represent all the values satisfying the equation as \(x=\frac{\pi}{3} + 2\pi n, n\in \mathbb{Z}\) and \(x=\frac{5\pi}{3} + 2\pi n, n\in \mathbb{Z}\). The \(2\pi\) represents a full rotation back to where we started and \(n\) is an integer (...-2,-1,0,1,2...) which represents how many rotations we’ve made (forwards or backwards).

Now for the other factor, \[\begin{aligned} \cos{x}+1 &= 0 \\ \cos{x} &= -1 \end{aligned}\] This is a quadrantal value, so I know that on \([0,2\pi]\) only \(x=\pi\) satisfies \(\cos{x}=-1\). Then, once again, we must include every full rotation from \(\pi\) as a value satisfying the equation. So we have that \(x=\pi+2\pi n, n \in \mathbb{Z}\) satisfies the equation. In other words, every odd value of \(\pi\).

So I have three equations representing solutions now corresponding to unbit circle values of \(\frac{\pi}{3}, \frac{5\pi}{3},\) and \(\pi\). So those three equations are valid answers but I want you to notice something. (draw unit circle) These solutions are equally spaced on the unit circle. They divide the unit circle into 3 equal pieces. The difference between \(\pi\) and \(\frac{\pi}{3}\) is \(\frac{2\pi}{3}\). The difference between \(\frac{5\pi}{3}\) and \(\pi\) is \(\frac{2\pi}{3}\). \(\frac{5\pi}{3} + \frac{2\pi}{3} =\frac{7\pi}{3}\), which is hitting the terminal side of \(\frac{\pi}{3}\) again. So all of our solutions are actually plus a multiple of \(\frac{2\pi}{3}\) from \(\frac{\pi}{3}\). So we can actually write one equation to represent all our solutions to this equation: \[x = \frac{\pi}{3} + \frac{2\pi n}{3} , n \in \mathbb{Z}\]

Solution: \(x=\dfrac{\pi}{6},\dfrac{3\pi}{4},\dfrac{5\pi}{6},\dfrac{7\pi}{6},\dfrac{7\pi}{4},\dfrac{11\pi}{6}\)

Solution Method: This equation is a cubic with two different trig functions, cotangent and cosecant. I do not want to mess with this in terms of two variables, so that’s where the Pythagorean identity comes in. The Pythagorean identity \(1+\cot^2{x}=\csc^2{x}\) relates cotangent and cosecant. So I’m going to use it to substitute \(\csc^2{x}\) in terms of cotangent; because the rest of my terms are cotangent. So we have \[\begin{aligned} \cot^3{x}-3\cot{x}&=4-\csc^2{x} \\ \cot^3{x}-3\cot{x}&=4-(1+\cot^2{x}) \\ \cot^3{x}-3\cot{x}&=4-1-\cot^2{x} \\ \cot^3{x}-3\cot{x}&=3-\cot^2{x} \end{aligned}\] Now I’m going to set all the terms equal to zero. \[\begin{aligned} \cot^3{x} + \cot^2{x} -3\cot{x} -3 &= 0 \end{aligned}\] So now I have this cubic in cotangent. I want to be able to factor this, so let’s see if I can do it by grouping. I can take a \(\cot^2{x}\) out of the first two terms and a -3 out of the last two terms. \[\begin{aligned} \cot^2{x}(\cot{x}+1) -3(\cot{x} +1) &= 0 \end{aligned}\] The factors match so, indeed, I can factor the cubic by grouping as \[\begin{aligned} (\cot^2{x}-3)(\cot{x}+1) &= 0 \end{aligned}\] So now I can take these two binomials and set them equal to 0 and solve for x.

First, \[\begin{aligned} \cot^2{x} - 3 &= 0 \\ \cot^2{x} & = 3 \\ \sqrt{\cot^2{x}} & = \pm \sqrt{3} \\ \cot{x} &= \pm \sqrt{3} \end{aligned}\] So now I need to think about what reference angle produces cotangent of \(\sqrt{3}\). Remember, \(\cot{x}= \frac{\cos{x}}{\sin{x}}\). It is not 0 or \(\pm\)1, so it will not be a quadrantal value. At \(\frac{\pi}{4}\) sine and cosine are both \(\frac{\sqrt{2}}{2}\), so \(\cot{\frac{\pi}{4}}=\frac{\cos{\frac{\pi}{4}}}{\sin{\frac{\pi}{4}}}=1\). So that leaves me with \(\frac{\pi}{6}\) or \(\frac{\pi}{3}\). At \(\frac{\pi}{6}\), \(\cot{\frac{\pi}{6}}=\frac{\cos{\frac{\pi}{6}}}{\sin{\frac{\pi}{6}}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}\), so \(\frac{\pi}{6}\) is the reference value satisfying \(\cot{x} = \sqrt{3}\). Now, the \(\pm\) means we are not restricting to any quadrants, so we include the \(\frac{\pi}{6}\) values for every quadrant. So the values in \([0,2\pi]\) satisfying the equation are \(x=\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}\).

Now for the second factor, \[\begin{aligned} \cot{x} &=1 \\ \cot{x} &= -1 \end{aligned}\] Cotangent is equal to \(\pm\) 1 when sine and cosine are the same, which we know occurs at multiples of \(\frac{\pi}{4}\). And cotangent is negative (ASTC, cotangent matches sign of tangent) in Q2 and Q3. So the values satisfying the equation at \(x=\frac{3\pi}{4}, \frac{7\pi}{4}\).

So then the values of \(x \in [0,2\pi]\) satisfying \(\cos^3{x}-3\cot{x}=4-\csc^2{x}\) are \(x=\frac{\pi}{6},\frac{3\pi}{4}, \frac{5\pi}{6},\frac{7\pi}{6},\frac{7\pi}{4},\frac{11\pi}{6}\).

Solution: \(x=0,\dfrac{\pi}{6}, \pi, \dfrac{11\pi}{6},2\pi\)

Solution Method: We want to turn this equation into something we can factor. Thinking about the identities I can apply to this equation, the ratio identity for tangent comes to mind. So let’s try substituting \(\tan{x}=\frac{\sin{x}}{\cos{x}}\). \[\begin{aligned} \sqrt{3}\tan{x}&=2\sin{x} \\ \sqrt{3}\left(\frac{\sin{x}}{\cos{x}}\right)&=2\sin{x} \end{aligned}\] Then I multiply both sides by cosine, and move everything to one side. \[\begin{aligned} \sqrt{3}\sin{x} = 2\sin{x}\cos{x} \\ \sqrt{3}\sin{x} - 2\sin{x}\cos{x} = 0 \end{aligned}\] This equation I can factor. I can factor out a \(\sin{x}\). \[\begin{aligned} \sin{x}(\sqrt{3} - 2\cos{x}) = 0 \end{aligned}\]So now I have two factors I can set equal to 0.

First, \[\sin{x} = 0\] This is a quadrantal value and the values satsifying sine equal to 0 are \(x=0,\pi,2\pi\). (We need to include both 0 and \(2\pi\) because both are included in the interval)

Second, \[\begin{aligned} \sqrt{3}-2\cos{x} &= 0 \\ \sqrt{3} &= 2\cos{x} \\ \frac{\sqrt{3}}{2} &= \cos{x} \end{aligned}\] The reference angle satisfying \(\cos{x}=\frac{\sqrt{3}}{2}\) is \(\frac{\pi}{6}\). And cosine is positive in Q1 and Q4. So the values in \([0,2\pi]\) satisfying the equation are \(x=\frac{\pi}{6}\) in Q1, and \(x=\frac{11\pi}{6}\) in Q4.

So all the values satisfying \(\sqrt{3}\tan{x}=2\sin{x}\) in \([0,2\pi]\) are \(x=0,\frac{\pi}{6}, \pi, \frac{11\pi}{6}, 2\pi\).

Solution: \(x=\dfrac{\pi}{6}, \dfrac{5\pi}{6},\dfrac{3\pi}{2}\)

Solution Method: Something I notice immediately here is I have a double angle. So let’s recall the double angle identities for cosine.

The first and third identity is still going to leave me with a cosine term opposite the sine term on the RHS. But the second identity will give me a quadratic in sine. So let’s substitute \(\cos{2x}=1-2\sin^2{x}\). \[\begin{aligned} \cos(2x)&=\sin{x} \\ 1-2\sin^2{x} &= \sin{x} \\ 0 &= 2\sin^2{x} + \sin{x} -1 \end{aligned}\] Now I’ll try to factor the quadratic of sine like I would the equation \(2x^2+x-1=0\). So I need two numbers that multiply to be -2 and sum to 1. So they must be 2 and -1.

Since the leading coefficient of the quadratic is not one, we will split \(\sin{x}\) into \(2\sin{x}\) and \(-\sin{x}\) and factor by grouping. \[\begin{aligned} 0 &= 2\sin^2{x} + \sin{x} -1 \\ 0 &= 2\sin^2{x} + 2\sin{x} - \sin{x} -1 \\ 0 &= 2\sin{x}(\sin{x} + 1) - (\sin{x}+1) \\ 0 &= (2\sin{x}-1)(\sin{x}+1) \end{aligned}\] So now I have two factors to set equal to zero.

First, \[\begin{aligned} 2\sin{x} - 1 &= 0 \\ 2\sin{x} &= 1 \\ \sin{x} &= \frac{1}{2} \end{aligned}\] The reference angle producing a sine value of one-half is \(\frac{\pi}{6}\). And sine is positive in Q1 and Q2. So the values of \(x \in [0,2\pi]\) satisfying the equation are \(x=\frac{\pi}{6}\) in Q1 and \(x=\frac{5\pi}{6}\) in Q2.

Second, \[\begin{aligned} \sin{x} + 1 &= 0 \\ \sin{x} &= -1 \end{aligned}\] -1 is a quadrantal value so the only value satisfying sine of -1 in \(x=\frac{3\pi}{2}\).

So the values \(x\in [0,2\pi]\) satisfying \(\cos(2x)=\sin{x}\) are \(x=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}.\)

Solution: \(x=\dfrac{5\pi}{4} + 2\pi n, n\in\mathbb{Z}\) and \(x=\dfrac{7\pi}{4} +2\pi n, n\in \mathbb{Z}\)

Solution Method: To solve this equation, we need to utilize the sum and difference identities for cosine.

Sum and Difference Identities \[\begin{align*} \cos{(A+B)}&=\cos{A}\cos{B}-\sin{A}\sin{B} \\ \cos{(A-B)}&=\cos{A}\cos{B}+\sin{A}\sin{B} \\ \end{align*}\]

With these identities, we can rewrite the equation as \[\begin{aligned} \cos\left(x+\frac{\pi}{4}\right)&-\cos\left(x-\frac{\pi}{4}\right)&=1 \\ \left( \cos{x}\cos{\frac{\pi}{4}} -\sin{x}\sin{\frac{\pi}{4}}\right) &- \left( \cos{x}\cos{\frac{\pi}{4}} +\sin{x}\sin{\frac{\pi}{4}}\right) &= 1 \\ \cancel{\cos{x}\cos{\frac{\pi}{4}}} -\sin{x}\sin{\frac{\pi}{4}} &- \cancel{\cos{x}\cos{\frac{\pi}{4}}} -\sin{x}\sin{\frac{\pi}{4}} &= 1 \\ &-2\sin{x}\sin{\frac{\pi}{4}} &= 1 \end{aligned}\] Since \(\frac{\pi}{4}\) is a special angle value I know that \(\sin{\frac{\pi}{4}}=\frac{\sqrt{2}}{2}\). So I substitute that in, \[\begin{aligned} -2\sin{x}\sin{\frac{\pi}{4}} &= 1 \\ -2\sin{x}\left(\frac{\sqrt{2}}{2}\right) &= 1 \\ -\sqrt{2}\sin{x} &= 1 \\ \sin{x} &= \frac{1}{-\sqrt{2}} \\ \sin{x} &= -\frac{\sqrt{2}}{2} \end{aligned}\] The reference angle producing a sine of \(\frac{\sqrt{2}}{2}\) is \(\frac{\pi}{4}\). Then (ASTC) sine is negative in Q3 and Q4. So the values between 0 and \(2\pi\) satisfying the equation are \(\frac{5\pi}{4}\) in Q3 and \(\frac{7\pi}{4}\) in Q4. To find all x satisfying the equation we have to consider every full rotation from \(\frac{5\pi}{4}\) and \(\frac{7\pi}{4}\) is also a solution. So we can write the full solution as \(x=\frac{5\pi}{4} + 2\pi n, n\in \mathbb{Z}\) and \(x=\frac{7\pi}{4} + 2\pi n, n\in \mathbb{Z}\). The \(2\pi\) represents a full rotation back to where we started and \(n\) is an integer (...-2,-1,0,1,2...) which represents how many rotations we’ve made (forwards or backwards).

 

Solution: \(x = \frac{\pi}{2}+ \pi n, n \in \mathbb{Z}\)

Solution Method: I can choose to start with either the LHS or the RHS for this equation, because I’m going to need to apply an identity to both.

For the RHS, I’ll apply the ratio identity for cotangent, \(\cot{x} = \frac{\cos{x}}{\sin{x}}\). \[\begin{aligned} \sin\left(\frac{\pi}{2}-x\right) &= \cot{x} \\ \sin\left(\frac{\pi}{2}-x\right) &= \frac{\cos{x}}{\sin{x}} \end{aligned}\]

So I now have terms of sine and cosine, but the sine terms have different arguments, \((\frac{\pi}{2}-x) \text{ and } x\). So I’ll use a cofunction identity to get \(\sin(\frac{\pi}{2}-x)\) in terms of a trig function with argument \(x\).

Cofunction Identities for Sine/Cosine

\[\begin{aligned} \cos{x}&=\sin{\left(\frac{\pi}{2}-x\right)}\\ \sin{x}&=\cos{\left(\frac{\pi}{2}-x\right)} \end{aligned}\]

So substituting \(\sin{\left(\frac{\pi}{2}-x\right)}=\cos{x}\), \[\begin{aligned} \sin\left(\frac{\pi}{2}-x\right) &= \frac{\cos{x}}{\sin{x}} \\ \cos{x} &= \frac{\cos{x}}{\sin{x}} \\ \sin{x}\cos{x} &= \cos{x} \end{aligned}\] Now it may be tempting to divide out a \(\cos{x}\) here but if we do that, we’ll actually lose a solution to the equation, the case where \(\cos{x}=0\). So instead, we’ll get everything on one side, equal to zero, and factor out a \(\cos{x}\). \[\begin{aligned} 0 &= \cos{x}-\sin{x}\cos{x} \\ 0 &= \cos{x}(1-\sin{x}) \end{aligned}\] So now I’ll set both these factors equal to 0 and solve for x.

\[\begin{aligned} \cos{x}=0 \end{aligned}\] The quadrantal angles satisfying cosine of 0 on the unit circle are \(\frac{\pi}{2}, \frac{3\pi}{2}\). But we were asked for all solutions. Trig functions are periodic, they repeat themselves every rotation. So every full rotation from \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) is another value satisfying the equation. Since every rotation is \(2\pi\) in measure, we can represent all the values satisfying the equation as \(x=\frac{\pi}{2} + 2\pi n, n\in \mathbb{Z}\) and \(x=\frac{3\pi}{2} + 2\pi n, n\in \mathbb{Z}\). The \(2\pi\) represents a full rotation back to where we started and \(n\) is an integer (...-2,-1,0,1,2...) which represents how many rotations we’ve made (forwards or backwards).

\[\begin{aligned} 1-\sin{x} = 0 \\ 1 = \sin{x} \end{aligned}\] 1 is another quadrantal value and the angle satisfying \(\sin{x}=1\) on the unit circle is \(\frac{\pi}{2}\). This value is already included as a solution of \(\cos{x}=0\), so we don’t need to do anything else.

Now, we could leave these two equations as they are, but I want to point something out. Notice that \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) are a \(\pi\) apart. So \(\frac{\pi}{2} + \pi = \frac{3\pi}{2}, + \pi = \frac{5\pi}{2}...\) So we can actually represent all our solutions in one equation: \(x = \frac{\pi}{2} + \pi n, n \in \mathbb{Z}.\)