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Solve : cos 3x. Cos^(3)x+sin 3x.sin^(3)x...

Solve : `cos 3x. Cos^(3)x+sin 3x.sin^(3)x=0`

Text Solution

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The correct Answer is:
`x=(n pi)/(2)pm (pi)/(4)`
`cos 3x.cos^(3)x + sin 3x.sin^(3)x=0`
`therefore (4cos^(3)x-3cos x) cos^(3)x + (3 sin x -4 sin^(3)x)sin^(3)x=0`
`therefore 4(cos^(6)x-sin^(6)x)-3(cos^(4)x-sin^(4)x)=0`
`therefore 4(cos^(6)x-sin^(6)x)-3(cos^(2)x-sin^(2)x)=0`
`therefore (cos^(2)x-sin^(2)x)(1-4cos^(2)x sin^(2)x)=0`
`rArr cos 2x=0` or `sin^(2) 2x = 1 = sin pi//2`
`therefore 2x=(2n+1)pi//2` or `2x = n pi pm pi//2`
`therefore x = (2n +1)pi//4` or `x = (n pi)/(2)pm(pi)/(4), n in Z`
`therefore x = (n pi)/(2) pm (pi)/(), n in Z`
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