Answer to Question #129076 in Trigonometry for Ali

Question

 Solve the equation for all x  0<x<360

  

Cos(2x) + ½sin(x) = 0 

Answer 1

We know that $cos(2x) = 1 - 2 sin^2(x)$ . Now equation looks like

$1−2sin^2(x) + 0.5 sin(x) = 0$

Multiply the whole equation by -2 (just for convenience):

$4 sin^2 (x) - sin(x) -2 =0$

Now we have a standard quadratic equation for sin(x)

$sin(x) = 1/8 (1 ±\sqrt{1+32}) =1/8 (1 ±\sqrt{33})$

For the first root ($sin x = 1/8(1+\sqrt{33})$ ) the solution is$x = (-1)^k arcsin(1/8(1+\sqrt{33})) + \pi k ≈ (-1)^k + \pi k$

k =0: $x_1 = 1 rad ≈ 57°$

k = 1: $x_2 = -1 + \pi = 2.14 rad ≈123°$

The same for the second root:

$x = (-1)^k arcsin(1/8(1-\sqrt{33})) + \pi k ≈ (-1)^k*(-0.635) + \pi k$

$k =1: x_3= (0.635 + \pi )rad≈3.78 rad ≈ 216°$

$k =2: x_4= (-0.635 + 2\pi) rad≈5.645 rad ≈ 323°$