Solve the trigonometric equations in the interval 02 Solve t

Solve the trigonometric equations in the interval [0,2?].

Solve the trigonometric equations in the interval [0, 2 pi]. 4sin^2(x/2) = 1

Solution

0x2

=>0x_/2

given 4sin²(x_/2) =1

=>sin²(x_/2) =1_/4

=>sin(x_/2) =-1_/2 and sin(x_/2) =1_/2

for sin(x_/2) =-1_/2 there is no solution in given interval

for sin(x_/2) =1_/2

(x_/2)=(_/6) ,(x_/2)=(5_/6)

=>x=(_/3) ,x=(5_/3)

so solutions to 4sin²(x_/2) =1 in the interval [0,2] are x=(_/3) ,(5_/3)