Given:
2csc^2x-cot^4x=-12csc2x−cot4x=−1
cscx=1/sinxcscx=1sinx
cotx=cosx/sinxcotx=cosxsinx
2(1/sinx)^2-(cosx/sinx)^4=-12(1sinx)2−(cosxsinx)4=−1
2/sin^2x-cos^4x/sin^4x=-12sin2x−cos4xsin4x=−1
Multiplying throughout by sin^4xsin4x
2sin^2x-cos^4x=-sin^4x2sin2x−cos4x=−sin4x
Transposing
2sin^2x=cos^4x+sin^4x2sin2x=cos4x+sin4x
cos^4x+sin^4x=(cos^2x+sin^2x)^2-2cos^2xsin^2xcos4x+sin4x=(cos2x+sin2x)2−2cos2xsin2x
2sin^2x=(cos^2x+sin^2x)^2-2cos^2xsin^2x2sin2x=(cos2x+sin2x)2−2cos2xsin2x
cos^2x+sin^2x=1cos2x+sin2x=1
2sin^2x=(1)^2-2cos^2xsin^2x2sin2x=(1)2−2cos2xsin2x
2sin^2x=1-2cos^2xsin^2x2sin2x=1−2cos2xsin2x
If
u=cos^2xu=cos2x
1-u=sin^2x1−u=sin2x
Substituting
2(1-u)=1-2u(1-u)2(1−u)=1−2u(1−u)
2-2u=1-2u+2u^22−2u=1−2u+2u2
2u^2-2u+2u+1-2=02u2−2u+2u+1−2=0
2u^2-1=02u2−1=0
2u^2=12u2=1
u^2=1/2u2=12
u=+-(1/sqrt2)u=±(1√2)
u=cos^2xu=cos2x
cos^2x=+-(1/sqrt2)cos2x=±(1√2)
Considering only positive value for real numbers
cos^2x=1/sqrt2cos2x=1√2
cosx=+-(1/sqrtsqrt2)cosx=±(1√√2)
theta=cos^-1(1/sqrtsqrt2), 2pi-cos^-1(1/sqrtsqrt2)θ=cos−1(1√√2),2π−cos−1(1√√2)