What is the general solution for 1+cos^2(x)=sin^4(x) ?

The general solution for 1+cos^2(x)=sin^4(x) is x= pi/2+2pin,x=(3pi)/2+2pin

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1+cos^2(x)=sin^4(x)

1 + cos^{2} (x) = sin^{4} (x)

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

1 + cos^{2} (x) = sin^{4} (x)

Subtract

sin^{4} (x)

from both sides

1 + cos^{2} (x) - sin^{4} (x) = 0

Rewrite using trig identities

2 - sin^{2} (x) - sin^{4} (x) = 0

Solve by substitution

sin (x) = 2 i, sin (x) = - 2 i, sin (x) = 1, sin (x) = - 1

sin (x) = 2 i :

No Solution

sin (x) = - 2 i :

No Solution

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (t) = cos (2 t)

arctan (x + 2) = arcsin (\frac{7}{25}) + arccos (\frac{4}{5})

sec (a) = \frac{38}{13}

tan (c) = 0.6538

(sec^{2} (a)) cos^{2} (a) = sin^{2} (a)

What is the general solution for 1+cos^2(x)=sin^4(x) ?
The general solution for 1+cos^2(x)=sin^4(x) is x= pi/2+2pin,x=(3pi)/2+2pin