What is the general solution for 2sin^2(θ)-sin(θ)-1=0,0<= θ<2pi ?

The general solution for 2sin^2(θ)-sin(θ)-1=0,0<= θ<2pi is θ= pi/2 ,θ=(7pi)/6 ,θ=(11pi)/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2sin^2(θ)-sin(θ)-1=0,0<= θ<2pi

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

θ = \frac{π}{2}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Solution steps

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

Solve by substitution

sin (θ) = 1, sin (θ) = - \frac{1}{2}

sin (θ) = 1, 0 \leq θ < 2 π : θ = \frac{π}{2}

sin (θ) = - \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Combine all the solutions

θ = \frac{π}{2}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

3 sin (x) = tan (x)

cos^{2} (θ) + cos (θ) = sin^{2} (θ)

cos (4 x) = \frac{3}{2}

2 sin^{2} (x) = 2 sin (x)

cos (2 x) + 3 sin (x) - 2 = 0

What is the general solution for 2sin^2(θ)-sin(θ)-1=0,0<= θ<2pi ?
The general solution for 2sin^2(θ)-sin(θ)-1=0,0<= θ<2pi is θ= pi/2 ,θ=(7pi)/6 ,θ=(11pi)/6