` sin 3 theta + cos 2 theta = 0` को हल कीजिए ।

Explanation

` sin 3 theta + cos 2 theta = 0 `
` rArr sin 3 theta = sin ((3pi)/2 - 2 theta)`
` rArr 3 theta = n pi + (-1)^(n)* ((3pi)/2 - 2theta)`
यदि ` n = 2k` तो ,
` 3 theta = 2kpi + (-1)^(2k)* ((3pi)/2 - 2theta)`
` rArr 3 theta = 2kpi + (3pi)/2 - 2theta [ :' (-1)^(2k) = 1] `
` rArr 5 theta = 2kx + (3pi)/2 `
` rArr theta = 1/5 (2kpi+(3pi)/2)`
यदि ` n= 2K +1`, तो
` 3 theta = (2k+1) pi +(-)^(2k+1)((3pi)/2 - 2theta)`
` rArr 3 theta = pi - ((3pi)/2 - 2theta) [ :' (-1)^(2k+1) = -1] `
` rArr theta = ( 2kpi - pi/2) `