SOLUTION: Solve the following equations given that 0° &#8804; &#952; < 360° : A) cos &#952; = -((sqrt3)/2) B) sin &#952;/2 - 1/2 = 0

Algebra ->  Trigonometry-basics -> SOLUTION: Solve the following equations given that 0° &#8804; &#952; < 360° : A) cos &#952; = -((sqrt3)/2) B) sin &#952;/2 - 1/2 = 0       Log On


   

Question 966763: Solve the following equations given that 0° ≤ θ < 360° :
A) cos θ = -((sqrt3)/2)
B) sin θ/2 - 1/2 = 0

Answer by ikleyn(52810) About Me  (Show Source):
You can put this solution on YOUR website!

A)  If  cos+%28theta%29 = -sqrt%283%29%2F2%29   then   theta = 150°   or   theta = 210°,   providing that  0° < theta < 360°.


B)  If  sin+%28theta%2F2%29 -1%2F2%29 = 0   then   sin+%28theta%2F2%29 = 1%2F2%29.

      It implies that  theta%2F2 = 30°  or  theta%2F2 = 150°.   Hence,  theta = 60°  and  theta = 2*150° = 300°  satisfies the given equation,  providing that  0° < theta < 360°.