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

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

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

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

No Solution for x \in R

Solution steps

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

Subtract

2

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

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

No Solution

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

No Solution

sin (x) = \frac{1 - 5}{2} : x = arcsin \frac{1 - 5}{2} + 2 πn, x = π + arcsin - \frac{1 - 5}{2} + 2 πn

sin (x) = - \frac{1 - 5}{2} : x = arcsin - \frac{1 - 5}{2} + 2 πn, x = π + arcsin \frac{1 - 5}{2} + 2 πn

Combine all the solutions

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

Since the equation is undefined for:

arcsin \frac{1 - 5}{2} + 2 πn, π + arcsin - \frac{1 - 5}{2} + 2 πn, arcsin - \frac{1 - 5}{2} + 2 πn, π + arcsin \frac{1 - 5}{2} + 2 πn

No Solution for x \in R

5 tan (x) = 15 cot (x)

x, arctan (x) = arctan (y)

2 cos (x) - 23 \cdot sin (x) = 8

\frac{( cot ( x ) - 3 )}{( 2 sin ( x ) + 1 )} = 0

sin^{2} (x) + 3 sin (x) - 1 = 0

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