What is the general solution for -sin(2x)+2cos(x)=0 ?

The general solution for -sin(2x)+2cos(x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-sin(2x)+2cos(x)=0

- sin (2 x) + 2 cos (x) = 0

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

- sin (2 x) + 2 cos (x) = 0

Rewrite using trig identities

2 cos (x) - 2 cos (x) sin (x) = 0

Factor

2 cos (x) - 2 cos (x) sin (x) : - 2 cos (x) (sin (x) - 1)

- 2 cos (x) (sin (x) - 1) = 0

Solving each part separately

cos (x) = 0 or sin (x) - 1 = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

sin (x) - 1 = 0 : x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

8 cos^{2} (x) - 4 = 0

tan (x) = - 5

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

sin (x) + sin (3 x) = 0

2 sin^{2} (θ) - 5 sin (θ) + 3 = 0

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