Solve `sqrt(3) cos theta-3 sin theta =4 sin 2 theta cos 3 theta`.
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We have `sqrt(3) cos theta -3 sin theta=2 (sin 5 theta- sin theta)` or `(sqrt(3)/2) cos theta-(1/2) sin theta= sin 5 theta` `rArr cos (theta+pi/6)= sin 5 theta = cos (pi/2- 5 theta)` `rArr theta + pi/6 = 2n pi pm (pi/2 - 5theta)` or `theta=(n pi)/3+pi/18 or theta=- (n pi)/2 +pi/6, AA n in Z`
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Solve : 3-2 cos theta -4 sin theta - cos 2theta+sin 2theta=0 .
cos theta-cos3 theta+cos5 theta-cos7 theta=4sin theta sin4 theta cos2 theta
If (sin theta+cos theta)/(sin theta-cos theta)=3 and theta is an acute angle, then the value of (3sin theta+4cos theta)/(8cos theta-3sin theta) is: यदि (sin theta+cos theta)/(sin theta-cos theta)=3 तो (3sin theta+4cos theta)/(8cos theta-3sin theta) का मान ज्ञात करें ।
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