What is the general solution for 4sin^2(x)+2sin(x)-2=0 ?

The general solution for 4sin^2(x)+2sin(x)-2=0 is x= pi/6+2pin,x=(5pi)/6+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 >

4sin^2(x)+2sin(x)-2=0

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

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

Solution steps

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

Solve by substitution

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

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

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

Combine all the solutions

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

x, cos (x y) = 0

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

tan (2 x - \frac{π}{6}) = 3

sin^{2} (x) + 6 sin (x) = 2 sin (x) - 3

sin (x) + cos (x) = \frac{1}{5}

What is the general solution for 4sin^2(x)+2sin(x)-2=0 ?
The general solution for 4sin^2(x)+2sin(x)-2=0 is x= pi/6+2pin,x=(5pi)/6+2pin,x=(3pi)/2+2pin