Question

Find all solutions of the equation in the interval [0, 2pi)
cosx (2 sinx+1)=0

Answer 1

x = pi/2 and 11pi/6

Explanation:

Given the trigonometry expression

cosx (2 sinx+1)=0

This can be written in two forms as;

cosx = 0 and 2sinx + 1 = 0

If cos x = 0

x = arccos0

x = 90degrees

x = pi/2

Similarly

2sinx + 1 = 0

2sinx = -1

sinx = -1/2

sinx =-0.5

x = aarcsin(-0.5)

x = -30

Since sin is negative in the third and 4th quadrant

x = 270 + 30

x = 330degrees

x = 330pi/180

x = 11pi/6