Question 496428
sin2theta + square root 2 cos theta=0
**
for 0 ≤ x ≤ 2π
sin2x +√2 cosx=0
2sinxcosx+√2cosx=0
cosx(2sinx+√2)=0
..
cosx=0
x=π/2 and 3π/2
..
2sinx+√2)=0
2sinx=-√2)
sinx=-√2/2
x=5&#960;/4 and 7&#960;/4 (in quadrants III and IV where sin<0)
ans:
x=&#960;/2, 3&#960;/2, 5&#960;/4, and 7&#960;/4