English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin^2(θ)-cos^2(θ)=1+cos(θ)

Solution

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

Solution

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

1 + cos (θ)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

= - cos (θ) - 2 cos^{2} (θ)

- cos (θ) - 2 cos^{2} (θ) = 0

Solve by substitution

- cos (θ) - 2 cos^{2} (θ) = 0

Let:

cos (θ) = u

- u - 2 u^{2} = 0

- u - 2 u^{2} = 0 : u = - \frac{1}{2}, u = 0

- u - 2 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} - u = 0

Solve with the quadratic formula

- 2 u^{2} - u = 0

Quadratic Equation Formula:

For

a = - 2, b = - 1, c = 0

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

(- 1)^{2} - 4 (- 2) \cdot 0 = 1

(- 1)^{2} - 4 (- 2) \cdot 0

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 2 \cdot 0

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 + 0

Add the numbers:

1 + 0 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 1 ) \pm 1}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 1}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 1}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 1}{2 ( - 2 )} : - \frac{1}{2}

\frac{- ( - 1 ) + 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 + 1}{- 2 \cdot 2}

Add the numbers:

1 + 1 = 2

= \frac{2}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- ( - 1 ) - 1}{2 ( - 2 )} : 0

\frac{- ( - 1 ) - 1}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1 - 1}{- 2 \cdot 2}

Subtract the numbers:

1 - 1 = 0

= \frac{0}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

The solutions to the quadratic equation are:

u = - \frac{1}{2}, u = 0

Substitute back

u = cos (θ)

cos (θ) = - \frac{1}{2}, cos (θ) = 0

cos (θ) = - \frac{1}{2}, cos (θ) = 0

cos (θ) = - \frac{1}{2} : θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

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}

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

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}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn, θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Popular Examples

cos(x)sin(x)-2cos(x)=0

cos (x) sin (x) - 2 cos (x) = 0

csc(x)=-2

csc (x) = - 2

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

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

tan(θ)=2sin(θ)

tan (θ) = 2 sin (θ)

cos(x)=(-1)/2

cos (x) = \frac{- 1}{2}