What is the general solution for 2sin(θ)=-sqrt(3),0<= θ<2pi ?

The general solution for 2sin(θ)=-sqrt(3),0<= θ<2pi is θ=(4pi)/3 ,θ=(5pi)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin(θ)=-sqrt(3),0<= θ<2pi

2 sin (θ) = - 3, 0 \leq θ < 2 π

θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

Solution steps

2 sin (θ) = - 3, 0 \leq θ < 2 π

Divide both sides by

2

sin (θ) = - \frac{3}{2}

General solutions for

sin (θ) = - \frac{3}{2}

θ = \frac{4 π}{3} + 2 πn, θ = \frac{5 π}{3} + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{4 π}{3}, θ = \frac{5 π}{3}

cos^{2} (φ) = sin^{2} (φ)

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

sin (x) = \frac{1}{10}

(sin (θ) - 1) (sin (θ) + 1) = 0

5 sec (θ) - 6 = 0

What is the general solution for 2sin(θ)=-sqrt(3),0<= θ<2pi ?
The general solution for 2sin(θ)=-sqrt(3),0<= θ<2pi is θ=(4pi)/3 ,θ=(5pi)/3