Question

Solve cosx.cos2x.cos3x = 1/4

Answer 1

cosxcos2xcos3x=1/4
or, (2cosxcos3x)(2cos2x)=1
or, (cos4x+cos2x)(2cos2x)=1
or, 2cos4xcos2x+(2cos²2x-1)=0
or, 2cos4xcos2x+cos4x=0
or, cos4x(2cos2x+1)=0
Either, cos4x=0
or, 4x=(2n+1)π/2
or, x=(2n+1)π/8
Or, 2cos2x+1=0
or, 2cos2x=-1
or, cos2x=-1/2
or, cos2x=cos(-π/3)
or, cos2x=cosπ/3 [neglecting the negative sign for cos]
or, 2x=2nπ⁺₋π/3
or, x=nπ⁺₋π/6
(⁺₋ means plus-minus)

Answer 2

Step-by-step explanation:

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