What is the general solution for 2sin^2(x)+sin(x)-3=0,6 ?

The general solution for 2sin^2(x)+sin(x)-3=0,6 is No Solution for x\in\mathbb{R}

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2sin^2(x)+sin(x)-3=0,6

2 sin^{2} (x) + sin (x) - 3 = 0, 6

No Solution for x \in R

Solution steps

2 sin^{2} (x) + sin (x) - 3 = 0, 6

Solve by substitution

sin (x) = 1, sin (x) = - \frac{3}{2}

sin (x) = 1, 6 :

No Solution

sin (x) = - \frac{3}{2}, 6 :

No Solution

Combine all the solutions

No Solution for x \in R

(1 - sin (2 x)) \cdot (1 - cos^{2} (x)) = 0

5 cos (x) - 10 sin (x) cos (x) = 0, 0 \leq x \leq \frac{π}{2}

∣ (211) ∣ ∣ (- 37 - 1) ∣ (cos (x)) = 0

u, x = 4 sinh (u)

tan (x) = \frac{25}{32}

What is the general solution for 2sin^2(x)+sin(x)-3=0,6 ?
The general solution for 2sin^2(x)+sin(x)-3=0,6 is No Solution for x\in\mathbb{R}