What is the general solution for 49sin(2β)=25,tan(β)<0 ?

The general solution for 49sin(2β)=25,tan(β)<0 is No Solution for β\in\mathbb{R}

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49sin(2β)=25,tan(β)<0

49 sin (2 β) = 25, tan (β) < 0

No Solution for β \in R

Solution steps

49 sin (2 β) = 25, tan (β) < 0

Divide both sides by

49

sin (2 β) = \frac{25}{49}

Apply trig inverse properties

2 β = arcsin (\frac{25}{49}) + 2 πn, 2 β = π - arcsin (\frac{25}{49}) + 2 πn

Solve

2 β = arcsin (\frac{25}{49}) + 2 πn : β = \frac{arcsin ( \frac{25}{49} )}{2} + πn

Solve

2 β = π - arcsin (\frac{25}{49}) + 2 πn : β = \frac{π}{2} - \frac{arcsin ( \frac{25}{49} )}{2} + πn

β = \frac{arcsin ( \frac{25}{49} )}{2} + πn, β = \frac{π}{2} - \frac{arcsin ( \frac{25}{49} )}{2} + πn

Solutions for the range

tan (β) < 0

No Solution for β \in R

cos (θ) = - 0.076

4 cos^{2} (θ) = 2, 0 \leq θ \leq 2 π

sin (θ) = \frac{21}{29}

3 tan (θ) + sec^{2} (θ) = 5

tan (t) = - \frac{3}{4}

What is the general solution for 49sin(2β)=25,tan(β)<0 ?
The general solution for 49sin(2β)=25,tan(β)<0 is No Solution for β\in\mathbb{R}