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

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

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sin^2(φ)+cos^2(φ)=2

sin^{2} (φ) + cos^{2} (φ) = 2

No Solution for φ \in R

Solution steps

sin^{2} (φ) + cos^{2} (φ) = 2

Subtract

cos^{2} (φ)

from both sides

sin^{2} (φ) = 2 - cos^{2} (φ)

Square both sides

(sin^{2} (φ))^{2} = (2 - cos^{2} (φ))^{2}

Subtract

(2 - cos^{2} (φ))^{2}

from both sides

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

Apply exponent rule:

a^{b} = a^{2} a^{b - 2}

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

Rewrite using trig identities

- 3 + 2 cos^{2} (φ) = 0

Solve by substitution

cos (φ) = \frac{3}{2}, cos (φ) = - \frac{3}{2}

cos (φ) = \frac{3}{2} :

No Solution

cos (φ) = - \frac{3}{2} :

No Solution

Combine all the solutions

No Solution

Verify solutions by plugging them into the original equation

No Solution for φ \in R

x, y^{2} (2 + sin (x)) = 1

2 = cos (x) + sin (x)

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

cos (2 x) - 5 cos (x) - 1 = 0

3 + 3 cos (θ) = 3

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