Question

Solve

tanx-3cotx=2tan3x(0<x<360)

cosx-sinx=cosy-siny

cos2xcotx+1=cos2x+cotx

Joshi sir comment

tanx - 3/tanx = 2[3tanx-tan3x]/[1-3tan2x]

so [tan2x - 3]/tanx = 2tanx[3-tan2x]/[1-3tan2x]

so [tan2x - 3][1-3tan2x] = 2tan2x[3-tan2x]

so [tan2x - 3][1-3tan2x] + 2tan2x[tan2x-3] = 0

so [tan2x - 3][1-tan2x] = 0

so tanx = 1, -1, √3, -√3

now check the values satisfying last eq.

then put these values in 2nd eq. to get answer.

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