1.

The number of solutions of the equation `cos6x+tan^2x+cos(6x)tan^2x=1`in the interval `[0,2pi]`is(a)`4`(b)` 5`(c) `6` (d)`7`

Answer» `cos6x+tan^2x+cos6x*tan^2x=1`
`cos6x[1+tan^2x]=1-tan^2x`
`cos6x=(1-tan^2x)/(1+tan^2x)`
`cos6x=cos2x`
`cos6x-cos2x=0`
`-2sin((6x+2x)/2)*sin((6x-2x)/2)=0`
`-2sin4xsin2x=0`
`sin4x=0`
`0<=x<=2pi`
`0<=4x<=8pi`
`4x=0,pi,2pi,3pi,4pi,5pi,6pi,7pi,8pi`
`x=0,pi/4,pi/2,3/4pi,pi,5/4pi,3/2pi,7/4pi,2pi`
`sin2x=0`
`0<=x<=2pi`
`0<=2x<=4pi`
`2x=0,pi,2pi,3pi,4pi`
`x=0,pi/2,pi/3/2pi,2pi`
`x=9`.


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