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

The general solution for 2cos^2(x)+cos(x)=1,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3 ,x=pi

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<= x<= 2pi

2 cos^{2} (x) + cos (x) = 1, 0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}, x = π

Solution steps

2 cos^{2} (x) + cos (x) = 1, 0 \leq x \leq 2 π

Solve by substitution

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

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

cos (x) = - 1, 0 \leq x \leq 2 π : x = π

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}, x = π

sin (\frac{θ}{2}) = \frac{7.83}{2 ( 4.35 )}

- 4 sin (c) - 2 = sin (c) + 1

cot^{2} (x) + 1 = 2

sin (x) = \frac{1}{3}, sec (y) = \frac{25}{24 , cos ( x + y )}

2 = 2 - π cos (t)

What is the general solution for 2cos^2(x)+cos(x)=1,0<= x<= 2pi ?
The general solution for 2cos^2(x)+cos(x)=1,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3 ,x=pi