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

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

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2+2sin(θ)=cos^2(θ)

Solution

2 + 2 sin (θ) = cos^{2} (θ)

Solution

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

2 + 2 sin (θ) = cos^{2} (θ)

Subtract

cos^{2} (θ)

from both sides

2 + 2 sin (θ) - cos^{2} (θ) = 0

Rewrite using trig identities

2 - cos^{2} (θ) + 2 sin (θ)

Use the Pythagorean identity:

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

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

= 2 - (1 - sin^{2} (θ)) + 2 sin (θ)

Simplify

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

2 - (1 - sin^{2} (θ)) + 2 sin (θ)

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

- (1 - sin^{2} (θ))

Distribute parentheses

= - (1) - (- sin^{2} (θ))

Apply minus-plus rules

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

= - 1 + sin^{2} (θ)

= 2 - 1 + sin^{2} (θ) + 2 sin (θ)

Subtract the numbers:

2 - 1 = 1

= sin^{2} (θ) + 2 sin (θ) + 1

= sin^{2} (θ) + 2 sin (θ) + 1

1 + sin^{2} (θ) + 2 sin (θ) = 0

Solve by substitution

1 + sin^{2} (θ) + 2 sin (θ) = 0

Let:

sin (θ) = u

1 + u^{2} + 2 u = 0

1 + u^{2} + 2 u = 0 : u = - 1

1 + u^{2} + 2 u = 0

Write in the standard form

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

u^{2} + 2 u + 1 = 0

Solve with the quadratic formula

u^{2} + 2 u + 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 2, c = 1

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

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

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

2^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- 2 \pm 0}{2 \cdot 1}

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

u = - 1

The solution to the quadratic equation is:

u = - 1

Substitute back

u = sin (θ)

sin (θ) = - 1

sin (θ) = - 1

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

sin (θ) = - 1

General solutions for

sin (θ) = - 1

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}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{3 π}{2} + 2 πn

Graph

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-19sin(θ)+19/2 =0

- 19 sin (θ) + \frac{19}{2} = 0

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picos(pix)+pisin(pix)=0

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144=100+6.93^2-2*10*6.93*cos(x)

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Frequently Asked Questions (FAQ)

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