What is the general solution for 18cos(x)+9sin(2x)=0 ?

The general solution for 18cos(x)+9sin(2x)=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 >

18cos(x)+9sin(2x)=0

18 cos (x) + 9 sin (2 x) = 0

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

Solution steps

18 cos (x) + 9 sin (2 x) = 0

Rewrite using trig identities

18 cos (x) + 18 cos (x) sin (x) = 0

Factor

18 cos (x) + 18 cos (x) sin (x) : 18 cos (x) (sin (x) + 1)

18 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{3 π}{2} + 2 πn

Combine all the solutions

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

sin (x) = \frac{5}{29}

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

sin (a) = \frac{3}{5}, cos (a)

cos (x) = cos (4 x)

11 sin (B) + 13 = 4 sin (B) + 8

What is the general solution for 18cos(x)+9sin(2x)=0 ?
The general solution for 18cos(x)+9sin(2x)=0 is x= pi/2+2pin,x=(3pi)/2+2pin