Solving Trig Equations Using Identities Exercise 5

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