Solve the following trigonometric equation in the interval 0 ≤ x ≤ 2π. Express your answers as exact values where applicable, otherwise round your answers to 2 decimal places.

Question

2sin^2 x − 3 sinx + 1 = 0

Answer 1

2sin^2 x − 3 sinx + 1 = 0

using shortcut, replacing sin(x) with x

2x^2-3x+1=0

(x-1)(x-1/2)=0

thus:

sin(x)-1=0 and sin(x)-1/2=0

sin(x) = 1 and sin(x)=1/2

arcsin(1)= 90 degree, pi/2 and arcsin(1/2)= 30 degree pi/6 and (5pi)/6

answer:

pi/2,pi/6,(5pi)/6