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

The general solution for 2sin^2(x)+sin^2(x)+cos^2(x)=1 is x=2pin,x=pi+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Move

1

to the right side

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

Subtract

1

from both sides

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

Simplify

2 sin^{2} (x) = 0

2 sin^{2} (x) = 0

Divide both sides by

2

2 sin^{2} (x) = 0

Divide both sides by

2

2 sin^{2} (x) = 0

Divide both sides by

2

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

Simplify

sin^{2} (x) = 0

sin^{2} (x) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin^{2} (x) + 3 cos (x) - 1 = 0

cos (\frac{3 x - 7}{2}) = 0

3 tan^{2} (x) = \frac{8}{sin ^{2} ( x )}

sin^{2} (x) + sin^{22} (x) + sin^{23} (x) = 1

5 sin (1.75 x) = 1.4

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