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

The general solution for -sin(x)+2cos^2(x)=1 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-sin(x)+2cos^2(x)=1

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

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

\frac{6}{π} - 4 sin (2 x) = 0

sin (t) = \frac{1}{5}

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

sin (θ) = \frac{- 1}{2}

(sin (a) + cos (a))^{2} = 1 + 2, a cos (a)

What is the general solution for -sin(x)+2cos^2(x)=1 ?
The general solution for -sin(x)+2cos^2(x)=1 is x=(3pi)/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin