Solving Trig Equations
Instructions
- This page includes exercises that you should attempt to solve yourself. You can check your answers and watch the videos explaining how to solve the exercises.
- When you are done, you can use the graph to pick another section or use the buttons to go to the next section.
Learning Objectives
- Using algebra techniques and knowledge of the unit circle to solve trigonometric equations
- Write an equation to represent all possible solutions of a trigonometric equation by accounting for coterminal angles
Exercises
Solution: \(x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}\).
Solution Method: This equation is a "quadratic in sine". It’s in the form of \(ax^2+bx+c=0\), but the variable is \(\sin{x}\). Since the equation is in a quadratic form, I can factor it like I would the quadratic equation \(2x^2+5x-3=0.\) To factorize \(2x^2+5x-3=0\), we need to find two numbers whose product is -6 and sum to 5. So those numbers are going to be -1 and 6 \(( 6(-1)=-6, 6-1 = 5)\). Since the leading coefficient of the quadratic isn’t 1, we’re going to split the 5 into 6 and -1, and factor by grouping. \[\begin{aligned} 2x^2+5x-3&=0 \\ 2x^2 + 6x - x -3 &= 0 \\ 2x(x +3) - (x+3) &= 0 \\ (2x-1)(x+3) &= 0 \end{aligned}\] So then the factorization of the quadratic in sine is \[\begin{aligned} 2\sin^2{x}+5\sin{x}-3 &= 0 \\ (2\sin{x} - 1) (\sin{x}+3) &= 0 \end{aligned}\] We need to find the \(\sin{x}\) values, then the x-values, that satisfy the equation, so we’re going to take each factor and set it to zero.
\[\begin{aligned} 2\sin{x}-1 &= 0 \\ 2\sin{x} &= 1 \\ \sin{x} &= \frac{1}{2} \end{aligned}\] So now we think back to special angle values and the unit circle. Sine of what reference angle produces \(\frac{1}{2}\)? \(\frac{\pi}{6}\). The half is positive so our acronym ASTC reminds us that sine is positive in Q1 and Q2. So the x values (in \([0,2\pi]\)) that satisfy \(\sin{x}=\frac{1}{2}\) are \(x = \frac{\pi}{6}, \frac{5\pi}{6}\).
\[\begin{aligned} \sin{x}+3 &= 0 \\ \sin{x} &= -3 \\ \end{aligned}\]
Think back to the graph of a sine function. It’s range is [-1,1]. So is it ever possible for \(\sin{x} = -3\)? Nope. So this equation yields no solution.
So the values of \(x\) in \([0,2\pi]\) satisfying \(2\sin^2{x}+5\sin{x}-3=0\) are \(x = \frac{\pi}{6}, \frac{5\pi}{6}\).
Solution: \(x=\dfrac{\pi}{4}, \dfrac{\pi}{2}, \dfrac{3\pi}{4},\dfrac{3\pi}{2}\)
Solution Method: A tempting first step here is to divide by a \(\cos{x}\) on both sides. But if we do that, we’ll be losing a solution to the equation! (Think about \(x^2=3x\). We can clearly see that \(x=0\) is a solution to the equation. But if we divide by x, we may think \(x=3\) is the only solution.)
So instead of dividing by \(\cos{x}\) as out first step, we’re going to move everything over to one side and set it equal to zero. \[\begin{aligned} 2\sin{x}\cos{x} &= \sqrt{2}\cos{x} \\ 2\sin{x}\cos{x} - \sqrt{2}\cos{x} &= 0 \end{aligned}\] Then I can factor out a \(\cos{x}\), \[\begin{aligned} 2\sin{x}\cos{x} - \sqrt{2}\cos{x} &= 0 \\ \cos{x}(2\sin{x} - \sqrt{2}) &= 0 \end{aligned}\] So I have two factors I can set equal to 0 and solve for x.
2 \[\begin{aligned} \cos{x} = 0 \end{aligned}\] Zero is a quadrantal value and thinking back to the unit circle, I know \(x=\frac{\pi}{2}, \frac{3\pi}{2}\) satisfy \(\cos{x}=0\). \[\begin{aligned} 2\sin{x}-\sqrt{2} &= 0 \\ 2\sin{x} = \sqrt{2} \\ \sin{x} = \frac{\sqrt{2}}{2} \end{aligned}\] The reference angle that produces sine equal to \(\frac{\sqrt{2}}{2}\) is \(\frac{\pi}{4}\). The sine value is also positive, so I’m in Q1 or Q2 (by ASTC). So the x values satisfying the equation are \(x=\frac{\pi}{4}, \frac{3\pi}{4}\).
So we have a total of four x-values satisfying \(2\sin{x}\cos{x} = \sqrt{2}\cos{x}\), \(x=\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \frac{3\pi}{2}.\)
Solution Method: This equation is a "quadratic in secant". If we just subtract the two over to set the LHS equal to zero, we have
\[
\sec^2{x}+\sec{x}-2=0
\]
So it's in the form of \(ax^2+bx+c=0\), but the variable is \(\sec{x}\). Since the equation is in a quadratic form, I can factor it like I would the quadratic equation \(x^2+x-2=0.\) The equation factors to
\begin{align*}
x^2+x-2&=0 \\
(x-1)(x+2) &=0
\end{align*}
So the quadratic in secant factors to
\begin{align*}
\sec^2{x}+\sec{x}-2&=0 \\
(\sec{x}-1)(\sec{x}+2) &= 0
\end{align*}
Then I have two factors to set equal to 0.
The first factor gives us
\begin{align*}
\sec{x}-1 &= 0 \\
\sec{x} &= 1
\end{align*}
Since secant is the reciprocal of cosine, if \(\sec{x}=1\) then
\begin{equation*}
\cos{x} = \frac{1}{1} = 1
\end{equation*}
1 is a quadrantal value, and thinking back to the unit circle I know that \(x=0, 2\pi\) satisfies \(\cos{x} = 1 \). However, the question asks us to find all values of x satisfying the equation, not just the ones in \([0,2\pi]\). Trig functions are periodic; they repeat themselves every \(2\pi\). We can see that in the graphs of sine, cosine, and tangent, and on the unit circle if we keep going around it. So if 0 and \(2\pi\) satisfy the equation, so will \(4\pi, 6\pi, 8\pi...\) and \(-2\pi,-4\pi,...\) See the pattern? They are all integer multiples of \(2\pi\). Once we find this pattern, instead of listing all of these values which are, in fact, infinite, we can write that the values \(x=2\pi n\), for all \(n \in \mathbb{Z}\) satisfy the equation.
Now for the second factor.
\begin{align*}
\sec{x}+2 &= 0 \\
\sec{x} &= -2
\end{align*}
Since secant is the reciprocal of cosine, if \(\sec{x}=-2\) then
\begin{equation*}
\cos{x} = \frac{1}{-2} = -\frac{1}{2}
\end{equation*}
The reference angle that produces a cosine value of \(\frac{1}{2}\) is \(\frac{\pi}{3}\). Then the quadrants in which cosine is negative are (thinking about ASTC) Q2 and Q3. So the angle values satisfying \(\cos{x} = -\frac{1}{2}\) between 0 and \(2\pi\) are \(x=\frac{2\pi}{3}, \frac{4\pi}{3}\). Once again we need to find all values of x satisfying the equation. So we must account for the fact that each full rotation of \(2\pi\) lands us back at an x value satisfying the equation. So \(x=\frac{2\pi}{3}+2\pi n, n\in\mathbb{Z}\) and \(x=\frac{4\pi}{3}+2\pi n, n\in\mathbb{Z}\) also satisfy the equation. The integer values multiplied by the \(2\pi\) represent full rotations.
Note: The equations representing all possible x values can be written in multiple ways. For example, \(x=\frac{2\pi}{3}+2\pi n, n\in\mathbb{Z}\) could also be written as \(x=\frac{1}{3}(2\pi+6\pi n), n\in\mathbb{Z}\). Or since \(\frac{4\pi}{3} = - \frac{2\pi}{3}, x=\frac{4\pi}{3}+2\pi n, n\in\mathbb{Z}\) could be written \(x=-\frac{2\pi}{3}+2\pi n, n\in\mathbb{Z}\)}
Solution: \(x=0,\pi,\frac{3\pi}{2}, 2\pi\)
Solution Method: While it might be tempting to divide everything by a \(\tan^2{x}\) here, if we do we’ll be losing a solution to the equation! (Think about \(x^2=3x\). We can clearly see that \(x=0\) is a solution to the equation. But if we divide by x, we may think \(x=3\) is the only solution.) So we cannot divide trig values out of the equation, but we can factor them out. For this equation, we have a common factor of \(\tan^2{x}\), so I can factor it out of the LHS. \[\begin{aligned} \tan^2{x}+\tan^2{x}\sin{x}&=0 \\ \tan^2{x}(1+\sin{x}) &= 0 \end{aligned}\] So I can set \(\tan^2{x}\) and \(1+\sin{x}\) equal to 0 separately.
\[\begin{aligned} \tan^2{x} &= 0 \\ \sqrt{\tan^2{x}} &= \sqrt{0} \\ \tan{x} &= 0 \\ \end{aligned}\] I know that \(\tan{x}\) is defined by the ratio identity \(\tan{x} = \frac{\sin{x}}{\cos{x}}\). So for \(\tan{x}=0\), \(\sin{x}\) must be 0. Cosine equaling zero will make tangent undefined, NOT zero! And \(\sin{x}=0\) is satisfied by the quadrantal angles \(x=0,\pi,2\pi\) (We must include both 0 and \(2\pi\) because the interval includes both as endpoints).
\[\begin{aligned} 1+\sin{x} &= 0 \\ \sin{x} &= -1 \end{aligned}\] Thinking back to the unit circle, I know that \(\sin{x} = -1\) is only satisfied in \([0,2\pi]\) by the quadrantal angle \(x=\frac{3\pi}{2}\).
So the values in \([0,2\pi]\) satisfying \(\tan^2{x}+\tan^2{x}\sin{x}=0\) are \(x=0,\pi,\frac{3\pi}{2}, 2\pi.\)