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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos^2(x)+1=3cos(x),0<= x<= 2pi

2 cos^{2} (x) + 1 = 3 cos (x), 0 \leq x \leq 2 π

x = 0, x = 2 π, x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

2 cos^{2} (x) + 1 = 3 cos (x), 0 \leq x \leq 2 π

Solve by substitution

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

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

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

Combine all the solutions

x = 0, x = 2 π, x = \frac{π}{3}, x = \frac{5 π}{3}

cos (x) = \frac{5}{9}

sin^{2} (x) = - \frac{1}{2}

4 tan^{2} (θ) = 11 tan (θ) + 3

6 tan^{2} (x) - 2 cos^{2} (x) = cos (2 x)

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

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