What is the general solution for d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x)) ?

The general solution for d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x)) is

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d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x))

d = \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 \cdot cos ^{2} ( x )}

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

Solution steps

d = \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )}

Switch sides

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} = d

Subtract

d

from both sides

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - d = 0

Simplify

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 cos ^{2} ( x )} - d : \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5 - 4 d cos ^{2} ( x )}{4 cos ^{2} ( x )}

\frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5 - 4 d cos ^{2} ( x )}{4 cos ^{2} ( x )} = 0

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

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2}, sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}, sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2}

sin (x) = \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} : x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} : x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} : x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

sin (x) = - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} : x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

Combine all the solutions

x = arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d + 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = arcsin - \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn, x = π + arcsin \frac{- 1 - 4 d - 16 d ^{2} + 24 d - 15}{2} + 2 πn

2 + cos^{2} (x) = 5 sin (x)

tan^{3} (3 x) - 2 sin^{3} (3 x) = 0

cot^{5} (x) = \frac{- 1}{( 3 )}

2 cos^{4} (x) cos (x) - cos^{5} (x) = 1

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

What is the general solution for d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x)) ?
The general solution for d=(sin^4(x)-cos^2(x)+5)/(4*cos^2(x)) is