What is the general solution for sin^2(x)-1+2cos(2x)-cos^2(x)=0 ?

The general solution for sin^2(x)-1+2cos(2x)-cos^2(x)=0 is x=pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = πn

Degrees

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

Solution steps

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Simplify

= - 1 + cos (2 x)

- 1 + cos (2 x) = 0

Move

1

to the right side

- 1 + cos (2 x) = 0

Add

1

to both sides

- 1 + cos (2 x) + 1 = 0 + 1

Simplify

cos (2 x) = 1

cos (2 x) = 1

General solutions for

cos (2 x) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 x = 0 + 2 πn

2 x = 0 + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

x = πn

cos^{2} (x) + 6 cos (x) + 5 = 0

sin (t) = - 0.9397

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

sinh (x) + 4 = 4 cosh (x)

sin (18 x) = - 1

What is the general solution for sin^2(x)-1+2cos(2x)-cos^2(x)=0 ?
The general solution for sin^2(x)-1+2cos(2x)-cos^2(x)=0 is x=pin