What is the general solution for sin(β)=-0,8(θ\in βvc)s=sec(β)-tan(β) ?

The general solution for sin(β)=-0,8(θ\in βvc)s=sec(β)-tan(β) is No Solution for β\in\mathbb{R}

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sin(β)=-0,8(θ\in βvc)s=sec(β)-tan(β)

sin (β) = - 0, 8 (θ \in β v c) s = sec (β) - tan (β)

No Solution for β \in R

Solution steps

sin (β) = - 0, 8 θ \in β v cs = sec (β) - tan (β)

sin (β) = 0

General solutions for

sin (β) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

β = 0 + 2 πn, β = π + 2 πn

β = 0 + 2 πn, β = π + 2 πn

Solve

β = 0 + 2 πn : β = 2 πn

β = 0 + 2 πn

0 + 2 πn = 2 πn

β = 2 πn

β = 2 πn, β = π + 2 πn

Solutions for the range

8 θ \in β v cs = sec (β) - tan (β)

No Solution for β \in R

cos (\frac{2 x - π}{17}) = 0

tan (X) cot (X) - tan (X) + 2 cot (X) = 0

sin (x) = - 0.3926

(sin (2 x) + 1) = 0, 0^{\circ} \leq x < 36 0^{\circ}

4 sin (x) + 2 = 2 sin (x)

What is the general solution for sin(β)=-0,8(θ\in βvc)s=sec(β)-tan(β) ?
The general solution for sin(β)=-0,8(θ\in βvc)s=sec(β)-tan(β) is No Solution for β\in\mathbb{R}