Solve the equation: `cos^2[pi/4(sinx+sqrt(2)cos^2x)]-tan^2[x+pi/4tan^2x]=1`

Explanation

`cos^2(pi/4(sinx+sqrt2cos^2x)]-tan^2(x+pi/4tan^2x)=1`

`-tan^2(x+pi/4tan^2x)=1-cos^2(pi/4(sinx+sqrt2cos^2x))`

`sin^2(pi/4(sinx+sqrt2cos^2x))+tan^2(x+pi/4tan^2x)=0`

`sin(pi/4(sinx+sqrt2cos^2x))=0`

`pi/4(sinx+sqrt2cos^2x)=npi`

`sinx+sqrt2(1-sin^2x)=4n`

`sinx+sqrt2-sqrt2sin^2x=0`

`sqrt2sin^2x-sinx-sqrt2=0`

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