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

The general solution for sin^2(x)+cos^4(x)=2 is No Solution for x\in\mathbb{R}

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

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

No Solution for x \in R

Solution steps

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

Subtract

2

from both sides

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = \frac{1 + 5}{2}, cos (x) = - \frac{1 + 5}{2}, cos (x) = \frac{1 - 5}{2}, cos (x) = - \frac{1 - 5}{2}

cos (x) = \frac{1 + 5}{2} :

No Solution

cos (x) = - \frac{1 + 5}{2} :

No Solution

cos (x) = \frac{1 - 5}{2} : x = arccos \frac{1 - 5}{2} + 2 πn, x = - arccos \frac{1 - 5}{2} + 2 πn

cos (x) = - \frac{1 - 5}{2} : x = arccos - \frac{1 - 5}{2} + 2 πn, x = - arccos - \frac{1 - 5}{2} + 2 πn

Combine all the solutions

x = arccos \frac{1 - 5}{2} + 2 πn, x = - arccos \frac{1 - 5}{2} + 2 πn, x = arccos - \frac{1 - 5}{2} + 2 πn, x = - arccos - \frac{1 - 5}{2} + 2 πn

Since the equation is undefined for:

arccos \frac{1 - 5}{2} + 2 πn, - arccos \frac{1 - 5}{2} + 2 πn, arccos - \frac{1 - 5}{2} + 2 πn, - arccos - \frac{1 - 5}{2} + 2 πn

No Solution for x \in R

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

cos (x) = 53

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

cos (x) = \frac{- 11}{12}

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

What is the general solution for sin^2(x)+cos^4(x)=2 ?
The general solution for sin^2(x)+cos^4(x)=2 is No Solution for x\in\mathbb{R}