Solve the equation for 0<x<360 2csc^2x-cot^4x=-1?

1 Answer
Mar 19, 2018

theta=cos^-1(1/sqrtsqrt2), 2pi-cos^-1(1/sqrtsqrt2)θ=cos1(12),2πcos1(12)

Explanation:

Given:
2csc^2x-cot^4x=-12csc2xcot4x=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=-12sin2xcos4xsin4x=1

Multiplying throughout by sin^4xsin4x

2sin^2x-cos^4x=-sin^4x2sin2xcos4x=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)22cos2xsin2x

2sin^2x=(cos^2x+sin^2x)^2-2cos^2xsin^2x2sin2x=(cos2x+sin2x)22cos2xsin2x

cos^2x+sin^2x=1cos2x+sin2x=1

2sin^2x=(1)^2-2cos^2xsin^2x2sin2x=(1)22cos2xsin2x

2sin^2x=1-2cos^2xsin^2x2sin2x=12cos2xsin2x

If
u=cos^2xu=cos2x
1-u=sin^2x1u=sin2x

Substituting

2(1-u)=1-2u(1-u)2(1u)=12u(1u)

2-2u=1-2u+2u^222u=12u+2u2

2u^2-2u+2u+1-2=02u22u+2u+12=0

2u^2-1=02u21=0

2u^2=12u2=1

u^2=1/2u2=12

u=+-(1/sqrt2)u=±(12)

u=cos^2xu=cos2x

cos^2x=+-(1/sqrt2)cos2x=±(12)

Considering only positive value for real numbers

cos^2x=1/sqrt2cos2x=12

cosx=+-(1/sqrtsqrt2)cosx=±(12)

theta=cos^-1(1/sqrtsqrt2), 2pi-cos^-1(1/sqrtsqrt2)θ=cos1(12),2πcos1(12)