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tags:
Cosine Fundamental Trigonometric Identity Sine Trigonometry Trigonometry Series

Solving Trig Equations Using Identities Exercise 5

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 \(0\leq x \leq 2\pi\) satisfying \(\cos(2x)=\sin({x}).\)

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.

Double Angle Identities for Cosine

\[\begin{align*}\cos{2x}&=\cos^2{x}-\sin^2{x} \\ \cos{2x}&=1-2\sin^2{x} \\ \cos{2x}&= 2\cos^2{x}-1 \end{align*}\]

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 \(x=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}.\)

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