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

The general solution for cos^2(θ)+2sin(θ)cos(θ)-sin^2(θ)=0 is θ=pin-pi/8 ,θ=pin+pi/2-pi/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos^2(θ)+2sin(θ)cos(θ)-sin^2(θ)=0

Solution

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

Solution

θ = πn - \frac{π}{8}, θ = πn + \frac{π}{2} - \frac{π}{8}

Degrees

θ = - 22. 5^{\circ} + 18 0^{\circ} n, θ = 67. 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Rewrite using trig identities

cos^{2} (θ) - sin^{2} (θ) + 2 cos (θ) sin (θ)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

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

Rewrite using trig identities

sin (2 θ) + cos (2 θ)

sin (2 θ) + cos (2 θ) = 2 sin (2 θ + \frac{π}{4})

sin (2 θ) + cos (2 θ)

Rewrite as

= 2 (\frac{1}{2} sin (2 θ) + \frac{1}{2} cos (2 θ))

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{1}{2}

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{1}{2}

= 2 (cos (\frac{π}{4}) sin (2 θ) + sin (\frac{π}{4}) cos (2 θ))

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= 2 sin (2 θ + \frac{π}{4})

= 2 sin (2 θ + \frac{π}{4})

2 sin (2 θ + \frac{π}{4}) = 0

Divide both sides by

2

2 sin (2 θ + \frac{π}{4}) = 0

Divide both sides by

2

\frac{2 sin ( 2 θ + \frac{π}{4} )}{2} = \frac{0}{2}

Simplify

sin (2 θ + \frac{π}{4}) = 0

sin (2 θ + \frac{π}{4}) = 0

General solutions for

sin (2 θ + \frac{π}{4}) = 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}

2 θ + \frac{π}{4} = 0 + 2 πn, 2 θ + \frac{π}{4} = π + 2 πn

2 θ + \frac{π}{4} = 0 + 2 πn, 2 θ + \frac{π}{4} = π + 2 πn

Solve

2 θ + \frac{π}{4} = 0 + 2 πn : θ = πn - \frac{π}{8}

2 θ + \frac{π}{4} = 0 + 2 πn

0 + 2 πn = 2 πn

2 θ + \frac{π}{4} = 2 πn

Move

\frac{π}{4}

to the right side

2 θ + \frac{π}{4} = 2 πn

Subtract

\frac{π}{4}

from both sides

2 θ + \frac{π}{4} - \frac{π}{4} = 2 πn - \frac{π}{4}

Simplify

2 θ = 2 πn - \frac{π}{4}

2 θ = 2 πn - \frac{π}{4}

Divide both sides by

2

2 θ = 2 πn - \frac{π}{4}

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 πn}{2} - \frac{\frac{π}{4}}{2}

Simplify

\frac{2 θ}{2} = \frac{2 πn}{2} - \frac{\frac{π}{4}}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{2 πn}{2} - \frac{\frac{π}{4}}{2} : πn - \frac{π}{8}

\frac{2 πn}{2} - \frac{\frac{π}{4}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= πn - \frac{π}{8}

θ = πn - \frac{π}{8}

θ = πn - \frac{π}{8}

θ = πn - \frac{π}{8}

Solve

2 θ + \frac{π}{4} = π + 2 πn : θ = πn + \frac{π}{2} - \frac{π}{8}

2 θ + \frac{π}{4} = π + 2 πn

Move

\frac{π}{4}

to the right side

2 θ + \frac{π}{4} = π + 2 πn

Subtract

\frac{π}{4}

from both sides

2 θ + \frac{π}{4} - \frac{π}{4} = π + 2 πn - \frac{π}{4}

Simplify

2 θ = π + 2 πn - \frac{π}{4}

2 θ = π + 2 πn - \frac{π}{4}

Divide both sides by

2

2 θ = π + 2 πn - \frac{π}{4}

Divide both sides by

2

\frac{2 θ}{2} = \frac{π}{2} + \frac{2 πn}{2} - \frac{\frac{π}{4}}{2}

Simplify

\frac{2 θ}{2} = \frac{π}{2} + \frac{2 πn}{2} - \frac{\frac{π}{4}}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{π}{2} + \frac{2 πn}{2} - \frac{\frac{π}{4}}{2} : πn + \frac{π}{2} - \frac{π}{8}

\frac{π}{2} + \frac{2 πn}{2} - \frac{\frac{π}{4}}{2}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{2} + πn - \frac{π}{8}

Group like terms

= πn + \frac{π}{2} - \frac{π}{8}

θ = πn + \frac{π}{2} - \frac{π}{8}

θ = πn + \frac{π}{2} - \frac{π}{8}

θ = πn + \frac{π}{2} - \frac{π}{8}

θ = πn - \frac{π}{8}, θ = πn + \frac{π}{2} - \frac{π}{8}

Graph

Popular Examples

sin(x)-cos(x)=0,0<= x<= 2pi

sin (x) - cos (x) = 0, 0 \leq x \leq 2 π

tan(x)+tan^2(x)=0

tan (x) + tan^{2} (x) = 0

(tan(x))/2 =1

\frac{tan ( x )}{2} = 1

sin(2θ)+2sin(θ)=0

sin (2 θ) + 2 sin (θ) = 0

tan(x/2+pi)+1=0

tan (\frac{x}{2} + π) + 1 = 0

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(θ)+2sin(θ)cos(θ)-sin^2(θ)=0 ?
The general solution for cos^2(θ)+2sin(θ)cos(θ)-sin^2(θ)=0 is θ=pin-pi/8 ,θ=pin+pi/2-pi/8