How do you solve #cos 2x = - sqrt3 / 2# using the double angle identity?

Question

Charles Autterson · Accepted Answer

Solve: #cos 2x = -sqrt3/2#

Ans: #+- 75^@ and +- 105^@# Solving by double angle identity: #cos 2a = 2cos^2 a - 1# #2cos^2 x - 1 = - sqrt3/2# #2cos^2 x = 1 - sqrt3/2 = (2 - sqrt3)/2# #cos^2 x = (2 - sqrt3)/4# cos x = +- sqrt(2 - sqrt3)/2 = +-0.517/2 = +- 0.258 a. cos x = 0.258 --> #x = +- 75^@# b. cos x = - 0.258 --> #x = +- 105^@# Check by calculator : #x = +- 75# --> #2x = +- 150# --> cos 2x = - 0.87 = -sqrt3/2. OK #x = +- 105# --> #2x = +-210# --> cos 2x = - 0.87. OK

Kai Anderson · Answer

undefined To solve $ \cos(2x) = -\frac{\sqrt{3}}{2} $ using the double angle identity, follow these steps:

1. Use the double angle identity $ \cos(2x) = 2\cos^2(x) - 1 $.

2. Substitute $ \cos(2x) $ with $ 2\cos^2(x) - 1 $ in the equation:
$$ 2\cos^2(x) - 1 = -\frac{\sqrt{3}}{2} $$

3. Add 1 to both sides of the equation:
$$ 2\cos^2(x) = 1 - \frac{\sqrt{3}}{2} $$

4. Simplify the right side:
$$ 2\cos^2(x) = \frac{2 - \sqrt{3}}{2} $$

5. Divide by 2:
$$ \cos^2(x) = \frac{2 - \sqrt{3}}{4} $$

6. Take the square root of both sides:
$$ \cos(x) = \pm \sqrt{\frac{2 - \sqrt{3}}{4}} $$

7. Simplify the square root:
$$ \cos(x) = \pm \frac{\sqrt{2 - \sqrt{3}}}{2} $$

8. Since $ \cos(x) = -\frac{\sqrt{3}}{2} $ lies in the third and fourth quadrants (where cosine is negative), the solution is:
$$ x = \pm \frac{5\pi}{6} + 2n\pi $$
$$ x = \pm \frac{7\pi}{6} + 2n\pi $$
where $ n $ is an integer.