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

The general solution for 2sin^2(x)+cos(x)-1=0,0<= x<= 2pi is x=(2pi)/3 ,x=(4pi)/3 ,x=0,x=2pi

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=0,0<= x<= 2pi

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

x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 0, x = 2 π

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{2 π}{3}, x = \frac{4 π}{3}

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

Combine all the solutions

x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 0, x = 2 π

sin (x) - tan (x) = cos (x) - 1

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

2 - 8 cos^{2} (x) = 0

2 tan (x) sec^{2} (x) = 0

sin (2 θ - \frac{π}{2}) = 1

What is the general solution for 2sin^2(x)+cos(x)-1=0,0<= x<= 2pi ?
The general solution for 2sin^2(x)+cos(x)-1=0,0<= x<= 2pi is x=(2pi)/3 ,x=(4pi)/3 ,x=0,x=2pi