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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = 0, x = 36 0^{\circ}, x = 12 0^{\circ}, x = 24 0^{\circ}

Solution steps

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

Solve by substitution

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

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

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

Combine all the solutions

x = 0, x = 36 0^{\circ}, x = 12 0^{\circ}, x = 24 0^{\circ}

θ, x = tan (θ)

sin (5 X) \cdot cos (5 X) = 1

100 + 60 sin (\frac{7 π}{3} x) = 100

sin^{2} (a) = 3 cos^{2} (a)

3 cot (x) + 3 = 0

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