What is the general solution for 2sin^2(θ)+5sin(θ)-12=0 ?

The general solution for 2sin^2(θ)+5sin(θ)-12=0 is No Solution for θ\in\mathbb{R}

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2sin^2(θ)+5sin(θ)-12=0

2 sin^{2} (θ) + 5 sin (θ) - 12 = 0

No Solution for θ \in R

Solution steps

2 sin^{2} (θ) + 5 sin (θ) - 12 = 0

Solve by substitution

sin (θ) = \frac{3}{2}, sin (θ) = - 4

sin (θ) = \frac{3}{2} :

No Solution

sin (θ) = - 4 :

No Solution

Combine all the solutions

No Solution for θ \in R

cos (2 θ) = \frac{3}{5}, 9 0^{\circ} \leq θ \leq 18 0^{\circ}

tan (θ) = \frac{1}{- 1}

sin^{2} (x) - 4 cos (x) - 4 = 0, 0 \leq x \leq 2 π

cos^{2} (θ) = 2 - 2 sin (θ)

x, sin (x) = 0.5

What is the general solution for 2sin^2(θ)+5sin(θ)-12=0 ?
The general solution for 2sin^2(θ)+5sin(θ)-12=0 is No Solution for θ\in\mathbb{R}