How do you solve cos 2x- sin^2 (x/2) + 3/4 = 0?

Aug 10, 2015

$C o s x = \frac{1}{2}$ and $\cos x = - \frac{3}{4}$

Explanation:

Step 1:
$\cos 2 x - S {\in}^{2} \left(\frac{x}{2}\right) + \frac{3}{4} = 0$
Use $\cos 2 x = {\cos}^{2} x - {\sin}^{2} x$

Step 2:
${\cos}^{2} x - {\sin}^{2} x - {\sin}^{2} \left(\frac{x}{2}\right) + \frac{3}{4} = 0$
Use ${\sin}^{2} x + {\cos}^{2} x = 1$

Step3:
$2 {\cos}^{2} x - 1 - {\sin}^{2} \left(\frac{x}{2}\right) + \frac{3}{4} = 0$
Use $\cos x = 1 - 2 {\sin}^{2} \left(\frac{x}{2}\right)$ (Double angle formula).

Step 4:
$2 {\cos}^{2} x - 1 - \frac{1}{2} + \frac{1}{2} \cos x + \frac{3}{4} = 0$
$2 {\cos}^{2} x + 2 \cos x - 3 = 0$

Multiply by 4 to get

$8 {\cos}^{x} + 2 \cos x - 3 = 0$

Step 5: Solve the quadratic equation to get

$\left(2 \cos - 1\right) \left(4 \cos x + 3\right) = 0$
$\cos x = \frac{1}{2}$ and $\cos x = - \frac{3}{4}$