What is the general solution for 1+2sin^2(x)=-5cos(x) ?

The general solution for 1+2sin^2(x)=-5cos(x) is x=(2pi)/3+2pin,x=(4pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1+2sin^2(x)=-5cos(x)

1 + 2 sin^{2} (x) = - 5 cos (x)

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Solution steps

1 + 2 sin^{2} (x) = - 5 cos (x)

Subtract

- 5 cos (x)

from both sides

1 + 2 sin^{2} (x) + 5 cos (x) = 0

Rewrite using trig identities

3 - 2 cos^{2} (x) + 5 cos (x) = 0

Solve by substitution

cos (x) = - \frac{1}{2}, cos (x) = 3

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

cos (x) = 3 :

No Solution

Combine all the solutions

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

sec^{2} (x) tan^{2} (x) + 3 sec^{2} (x) - 2 tan^{2} (x) = 3

2 cos (x) - 3 = 0

cot (x) + 6 sin (x) - 2 cos (x) = 3

cos (2 θ) = cos^{2} (θ)

cos (a) = \frac{12}{13}

What is the general solution for 1+2sin^2(x)=-5cos(x) ?
The general solution for 1+2sin^2(x)=-5cos(x) is x=(2pi)/3+2pin,x=(4pi)/3+2pin