What is the general solution for -2sin^2(x)-cos(x)=-1,0<= x<360 ?

The general solution for -2sin^2(x)-cos(x)=-1,0<= x<360 is x=0,x=120,x=240

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-2sin^2(x)-cos(x)=-1,0<= x<360

- 2 sin^{2} (x) - cos (x) = - 1, 0^{\circ} \leq x < 36 0^{\circ}

x = 0, x = 12 0^{\circ}, x = 24 0^{\circ}

Solution steps

- 2 sin^{2} (x) - cos (x) = - 1, 0^{\circ} \leq x < 36 0^{\circ}

Subtract

- 1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

cos (x) = 1, cos (x) = - \frac{1}{2}

cos (x) = 1, 0 \leq x < 36 0^{\circ} : x = 0

cos (x) = - \frac{1}{2}, 0 \leq x < 36 0^{\circ} : x = 12 0^{\circ}, x = 24 0^{\circ}

Combine all the solutions

x = 0, x = 12 0^{\circ}, x = 24 0^{\circ}

cos^{2} (θ) - 3 cos (θ) + 2 = sin^{2} (θ)

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

2 + 2 sin (θ) = - 4 sin (θ)

cos (θ) = \frac{( 5 )}{( 13 )}

2 sin (θ) + 3 = - csc (θ)

What is the general solution for -2sin^2(x)-cos(x)=-1,0<= x<360 ?
The general solution for -2sin^2(x)-cos(x)=-1,0<= x<360 is x=0,x=120,x=240