Answer:
[tex]\displaystyle x=\left\{\arcsin\left(\frac{-1+\sqrt{3}}{2}\right), \pi-\arcsin\left(\frac{-1+\sqrt{3}}{2}\right)\right\}[/tex]
Or their approximations:
[tex]x\approx \left\{0.375, 2.767\right\}[/tex]
Step-by-step explanation:
We are given:
[tex]\cos(2x)=2\sin(x)[/tex]
And we want to find the solution in [0, 2π).
Recall the double-angle identities for cosine:
[tex]\begin{aligned} \cos(2x)&=\cos^2(x)-\sin^2(x) \\&=2\cos^2(x)-1\\&=1-2\sin^2(x)\end{aligned}[/tex]
We will use the third version. Hence:
[tex]1-2\sin^2(x)=2\sin(x)[/tex]
Move all terms to one side:
[tex]-2\sin^2(x)-2\sin(x)+1=0[/tex]
This is now in quadratic form. For simplicity, let u = sin(x):
[tex]-2u^2-2u+1=0[/tex]
Solve for u. Simplify:
[tex]2u^2+2u-1=0[/tex]
By the quadratic formula:
[tex]\displaystyle u=\frac{-(2)\pm\sqrt{(2)^2-4(2)(-1)}}{2(2)}}[/tex]
Evaluate:
[tex]\displaystyle u=\frac{-1+\sqrt{3}}{2}\approx 0.366\text{ and } u=\frac{-1-\sqrt{3}}{2}\approx-1.366[/tex]
Note that the second solution is > -1. Hence, we will disregard it. (The range of sine is only -1 ≤ y ≤ 1.)
Back-substitute:
[tex]\displaystyle \sin(x)=\frac{-1+\sqrt{3}}{2}[/tex]
Since it is approximately 0.366, it will occur twice (once in QI and again in QII. This is because sine is positive only in those two quadrants). Using a calculator:
[tex]\displaystyle x_1=\arcsin\left(\frac{-1+\sqrt{3}}{2}\right)\approx0.375[/tex]
Using reference angles, the other solution is:
[tex]\displaystyle x_2=\pi -x_1=\pi -\arcsin\left(\frac{-1+\sqrt{3}}{2}\right) \approx2.767[/tex]
