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

The general solution for 2sin^2(x)+4cos^2(x)=1 is No Solution for x\in\mathbb{R}

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

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

No Solution for x \in R

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = \frac{3}{2}, sin (x) = - \frac{3}{2}

sin (x) = \frac{3}{2} :

No Solution

sin (x) = - \frac{3}{2} :

No Solution

Combine all the solutions

No Solution for x \in R

- sin (θ) = \frac{n}{2} - 3

3 sin (θ) + 2 sin^{2} (θ) = - 1, 0 \leq θ < 2 π

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

cos (x) = \frac{9}{14}

sin (θ) = \frac{7}{9}

What is the general solution for 2sin^2(x)+4cos^2(x)=1 ?
The general solution for 2sin^2(x)+4cos^2(x)=1 is No Solution for x\in\mathbb{R}