Please Help! Trigonometry Problem:
find the solution in 0< x < 2π
cos 2x = 2sinx

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]

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