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Cosine Factoring Fundamental Trigonometric Identity Quadratic Sine Trigonometry Trigonometry Series

Solving Trig Equations Using Identities Exercise 2

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Find all the solutions to  \(2\sin^2{x}=1+\cos{x}.\)

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: \[x = \frac{\pi}{3} + \frac{2\pi n}{3} , n \in \mathbb{Z}\]

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