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

The general solution for -2(cos(2x)+sin(x))=0 is x= pi/2+2pin,x=(7pi)/6+2pin,x=(11pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solution steps

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

Rewrite using trig identities

- (1 + sin (x) - 2 sin^{2} (x)) \cdot 2 = 0

Solve by substitution

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

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

6 tan (x) - 5 csc (x) = 0

2 (19) tan (\frac{θ}{2}) = 6.3

3 csc (x) - 6 = 0, 0^{\circ} \leq x \leq 36 0^{\circ}

tan (\frac{x}{2} + \frac{π}{3}) = - 1

What is the general solution for -2(cos(2x)+sin(x))=0 ?
The general solution for -2(cos(2x)+sin(x))=0 is x= pi/2+2pin,x=(7pi)/6+2pin,x=(11pi)/6+2pin