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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

- cos^{2} (x)

from both sides

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

Rewrite using trig identities

sin^{2} (x) + sin (x) + cos^{2} (x)

Use the Pythagorean identity:

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

= sin (x) + 1

sin (x) + 1 = 0

Move

1

to the right side

sin (x) + 1 = 0

Subtract

1

from both sides

sin (x) + 1 - 1 = 0 - 1

Simplify

sin (x) = - 1

sin (x) = - 1

General solutions for

sin (x) = - 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

arctan (x) = arccos (\frac{3}{4})

tan^{2} (θ) + 7 tan (θ) + 9 = 0

x, cos (x) \cdot cos (y) = 0

3 sec (θ) = 6

cot (x) = csc (x) + 3

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