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

The general solution for cos(2θ)=sin(θ),0<= θ<2pi is θ=(3pi)/2 ,θ= pi/6 ,θ=(5pi)/6

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<= θ<2pi

cos (2 θ) = sin (θ), 0 \leq θ < 2 π

θ = \frac{3 π}{2}, θ = \frac{π}{6}, θ = \frac{5 π}{6}

Solution steps

cos (2 θ) = sin (θ), 0 \leq θ < 2 π

Subtract

sin (θ)

from both sides

cos (2 θ) - sin (θ) = 0

Rewrite using trig identities

1 - sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = - 1, sin (θ) = \frac{1}{2}

sin (θ) = - 1, 0 \leq θ < 2 π : θ = \frac{3 π}{2}

sin (θ) = \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{π}{6}, θ = \frac{5 π}{6}

Combine all the solutions

θ = \frac{3 π}{2}, θ = \frac{π}{6}, θ = \frac{5 π}{6}

sin (x) (1 + cos (x)) = sin (x) + cos (x)

9 0^{2} = 6 0^{2} + 7 0^{2} - 2 (60) (70) (cos (x))

sin^{2} (x) + 3 sin (x) + 7 = 5

cos (2 x) = - 0.6, - 18 0^{\circ} \leq x \leq 18 0^{\circ}

23 cos (\frac{θ}{2}) = - 3

What is the general solution for cos(2θ)=sin(θ),0<= θ<2pi ?
The general solution for cos(2θ)=sin(θ),0<= θ<2pi is θ=(3pi)/2 ,θ= pi/6 ,θ=(5pi)/6