What is the general solution for cos(2θ)+sin(θ)=0,0<= θ<2pi ?

The general solution for cos(2θ)+sin(θ)=0,0<= θ<2pi is θ=(7pi)/6 ,θ=(11pi)/6 ,θ= pi/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(2θ)+sin(θ)=0,0<= θ<2pi

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

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

Solution steps

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

Rewrite using trig identities

1 + sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - \frac{1}{2}, sin (θ) = 1

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

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

Combine all the solutions

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

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

\frac{sin ( x )}{1} = \frac{3}{5}

x, e^{y} = tan (x + \frac{π}{3})

cos (x) = (sin (x))^{2} - 1

3 cos^{2} (x) - sin (x) - 1 = 0

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