What is the general solution for 2sin^2(x)+1=0 ?

The general solution for 2sin^2(x)+1=0 is No Solution for x\in\mathbb{R}

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2sin^2(x)+1=0

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

No Solution for x \in R

Solution steps

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

Solve by substitution

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

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

No Solution

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

No Solution

Combine all the solutions

No Solution for x \in R

sin (x + \frac{π}{4}) - sin (x - \frac{π}{4}) = 0

4 tan (x) - tan^{2} (x) = 1

x, arcsin (y) = arcsin (x) + \frac{( π )}{4}

18 sin^{2} (x) - 9 sin (x) = 0

3 sin^{2} (θ) + 6 sin (θ) - 11 = 7 sin (θ) - 9

What is the general solution for 2sin^2(x)+1=0 ?
The general solution for 2sin^2(x)+1=0 is No Solution for x\in\mathbb{R}