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

The general solution for sin^2(θ)=2cos(θ)+1 is θ= pi/2+2pin,θ=(3pi)/2+2pin

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sin^2(θ)=2cos(θ)+1

Solution

sin^{2} (θ) = 2 cos (θ) + 1

Solution

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

Degrees

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

Solution steps

sin^{2} (θ) = 2 cos (θ) + 1

Subtract

2 cos (θ) + 1

from both sides

sin^{2} (θ) - 2 cos (θ) - 1 = 0

Rewrite using trig identities

- 1 + sin^{2} (θ) - 2 cos (θ)

Use the Pythagorean identity:

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

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

= - 2 cos (θ) - cos^{2} (θ)

- cos^{2} (θ) - 2 cos (θ) = 0

Solve by substitution

- cos^{2} (θ) - 2 cos (θ) = 0

Let:

cos (θ) = u

- u^{2} - 2 u = 0

- u^{2} - 2 u = 0 : u = - 2, u = 0

- u^{2} - 2 u = 0

Solve with the quadratic formula

- u^{2} - 2 u = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 2^{2} + 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} + 0

2^{2} + 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Remove parentheses:

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

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

Add the numbers:

2 + 2 = 4

= \frac{4}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{- 2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

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

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

Remove parentheses:

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

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

Subtract the numbers:

2 - 2 = 0

= \frac{0}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{- 2}

Apply the fraction rule:

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

= - \frac{0}{2}

Apply rule

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

= - 0

= 0

The solutions to the quadratic equation are:

u = - 2, u = 0

Substitute back

u = cos (θ)

cos (θ) = - 2, cos (θ) = 0

cos (θ) = - 2, cos (θ) = 0

cos (θ) = - 2 :

No Solution

cos (θ) = - 2

- 1 \leq cos (x) \leq 1

No Solution

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

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(θ)=2cos(θ)+1 ?
The general solution for sin^2(θ)=2cos(θ)+1 is θ= pi/2+2pin,θ=(3pi)/2+2pin