The number of integral values of k for which the equation sinx(3ā2sin2x)+cosx(3ā2cos2x)=k possesses a solution, is
A
4
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B
5
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C
6
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D
7
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Solution
The correct option is B5 sinx(3−2sin2x)+cosx(3−2cos2x)=k Let sinx=s&cosx=c ⇒3(s+c)−2(s3+c3)=k⇒3(s+c)−2[(s+c)(s2+c2−sc)]=k⇒(s+c)[3−2(1−sc)]=k⇒(s+c)(1+2sc)=k⇒(s+c)3=k⇒k∈[−2√2,2√2]