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

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

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cos^4(x)=(sin^2(x)-1)/4

cos^{4} (x) = \frac{sin ^{2} ( x ) - 1}{4}

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

Solution steps

cos^{4} (x) = \frac{sin ^{2} ( x ) - 1}{4}

Subtract

\frac{sin ^{2} ( x ) - 1}{4}

from both sides

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

Simplify

cos^{4} (x) - \frac{sin ^{2} ( x ) - 1}{4} : \frac{4 cos ^{4} ( x ) - sin ^{2} ( x ) + 1}{4}

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

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = 0, cos (x) = i \frac{1}{2}, cos (x) = - i \frac{1}{2}

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = i \frac{1}{2} :

No Solution

cos (x) = - i \frac{1}{2} :

No Solution

Combine all the solutions

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

2 \cdot sin (x) + 1 = 0

5 cos^{2} (2 x) + 4 cos^{2} (x) - 5 = 0

sin (3 x + 10) = cos (x + 2 4^{\circ})

tan (x^{2}) + 1 = 0

\frac{( 1 + cos ^{2} ( a ) )}{sin ^{2} ( a )} = \frac{5}{3}

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