en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sqrt(2)sin^2(θ)+sin(θ)=0,0<= θ<2pi

Solution

2 sin^{2} (θ) + sin (θ) = 0, 0 \leq θ < 2 π

Solution

θ = 0, θ = π, θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

+1

Solution steps

2 sin^{2} (θ) + sin (θ) = 0, 0 \leq θ < 2 π

Solve by substitution

sin (θ) = 0, sin (θ) = - \frac{2}{2}

sin (θ) = 0, 0 \leq θ < 2 π : θ = 0, θ = π

sin (θ) = - \frac{2}{2}, 0 \leq θ < 2 π : θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

Combine all the solutions

θ = 0, θ = π, θ = \frac{5 π}{4}, θ = \frac{7 π}{4}

Popular Examples

cos(x)(2cos(x)+1)=3

cos (x) (2 cos (x) + 1) = 3

sqrt(3)+cot(x)=0

3 + cot (x) = 0

sin (θ) = - \frac{5}{8}

(sin(θ))/(cos(θ))=1

\frac{sin ( θ )}{cos ( θ )} = 1

tan(θ+26.565)=(12.5^2)/(9.81(60))

tan (θ + 26.56 5^{\circ}) = \frac{12. 5 ^{2}}{9.81 ( 60 )}

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)sin^2(θ)+sin(θ)=0,0<= θ<2pi ?
The general solution for sqrt(2)sin^2(θ)+sin(θ)=0,0<= θ<2pi is θ=0,θ=pi,θ=(5pi)/4 ,θ=(7pi)/4