What is the general solution for sin(θ)=-7/25 ,sin(2θ) ?

The general solution for sin(θ)=-7/25 ,sin(2θ) is No Solution for θ\in\mathbb{R}

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sin(θ)=-7/25 ,sin(2θ)

sin (θ) = - \frac{7}{25}, sin (2 θ)

No Solution for θ \in R

Solution steps

sin (θ) = - \frac{7}{25}, sin (2 θ)

Apply trig inverse properties

sin (θ) = - \frac{7}{25}

General solutions for

sin (θ) = - \frac{7}{25}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (- \frac{7}{25}) + 2 πn, θ = π + arcsin (\frac{7}{25}) + 2 πn

θ = arcsin (- \frac{7}{25}) + 2 πn, θ = π + arcsin (\frac{7}{25}) + 2 πn

Solutions for the range

sin (2 θ)

No Solution for θ \in R

cos (θ) = \frac{9}{12}

csc (x) + cot (x) = - 1, 0 \leq x \leq 2 π

sin (x) + cos (x) = \frac{5}{7}, 9 0^{\circ} > x > 0^{\circ}

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

2 cos (4 5^{\circ} - x) = 1

What is the general solution for sin(θ)=-7/25 ,sin(2θ) ?
The general solution for sin(θ)=-7/25 ,sin(2θ) is No Solution for θ\in\mathbb{R}