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

The general solution for 2cos^2(θ)-sin(θ)=2 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=2pin,θ=pi+2pin

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2cos^2(θ)-sin(θ)=2

Solution

2 cos^{2} (θ) - sin (θ) = 2

Solution

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = 2 πn, θ = π + 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (θ) - sin (θ) = 2

Subtract

2

from both sides

2 cos^{2} (θ) - sin (θ) - 2 = 0

Rewrite using trig identities

- 2 - sin (θ) + 2 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= - 2 - sin (θ) + 2 (1 - sin^{2} (θ))

Simplify

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

- 2 - sin (θ) + 2 (1 - sin^{2} (θ))

Expand

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

2 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (θ)

= 2 \cdot 1 - 2 sin^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (θ)

= - 2 - sin (θ) + 2 - 2 sin^{2} (θ)

Simplify

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

- 2 - sin (θ) + 2 - 2 sin^{2} (θ)

Group like terms

= - sin (θ) - 2 sin^{2} (θ) - 2 + 2

- 2 + 2 = 0

= - 2 sin^{2} (θ) - sin (θ)

= - 2 sin^{2} (θ) - sin (θ)

= - 2 sin^{2} (θ) - sin (θ)

- sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

- sin (θ) - 2 sin^{2} (θ) = 0

Let:

sin (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 = sin (θ)

sin (θ) = - \frac{1}{2}, sin (θ) = 0

sin (θ) = - \frac{1}{2}, sin (θ) = 0

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = - \frac{1}{2}

General solutions for

sin (θ) = - \frac{1}{2}

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{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = 2 πn, θ = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(θ)-sin(θ)=2 ?
The general solution for 2cos^2(θ)-sin(θ)=2 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ=2pin,θ=pi+2pin