What is the general solution for 2cos^2(θ)-1=sin(θ),0<= θ<= 180 ?

Q: What is the general solution for 2cos^2(θ)-1=sin(θ),0<= θ<= 180 ?

The general solution for 2cos^2(θ)-1=sin(θ),0<= θ<= 180 is θ=30,θ=150

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2cos^2(θ)-1=sin(θ),0<= θ<= 180

Solution

2 cos^{2} (θ) - 1 = sin (θ), 0^{\circ} \leq θ \leq 18 0^{\circ}

Solution

θ = 3 0^{\circ}, θ = 15 0^{\circ}

Radians

θ = \frac{π}{6}, θ = \frac{5 π}{6}

Solution steps

2 cos^{2} (θ) - 1 = sin (θ), 0^{\circ} \leq θ \leq 18 0^{\circ}

Subtract

sin (θ)

from both sides

2 cos^{2} (θ) - 1 - sin (θ) = 0

Rewrite using trig identities

- 1 - sin (θ) + 2 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= - 1 - sin (θ) + 2 (1 - sin^{2} (θ))

Simplify

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

- 1 - 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)

= - 1 - sin (θ) + 2 - 2 sin^{2} (θ)

Simplify

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

- 1 - sin (θ) + 2 - 2 sin^{2} (θ)

Group like terms

= - sin (θ) - 2 sin^{2} (θ) - 1 + 2

Add/Subtract the numbers:

- 1 + 2 = 1

= - 2 sin^{2} (θ) - sin (θ) + 1

= - 2 sin^{2} (θ) - sin (θ) + 1

= - 2 sin^{2} (θ) - sin (θ) + 1

1 - sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

1 - sin (θ) - 2 sin^{2} (θ) = 0

Let:

sin (θ) = u

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

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 (- 2) \cdot 1 = 3

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

Apply rule

- (- a) = a

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

(- 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 1 = 8

4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 3}{2 ( - 2 )}, u_{2} = \frac{- ( - 1 ) - 3}{2 ( - 2 )}

u = \frac{- ( - 1 ) + 3}{2 ( - 2 )} : - 1

\frac{- ( - 1 ) + 3}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 + 3}{- 2 \cdot 2}

Add the numbers:

1 + 3 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{- 4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 1 ) - 3}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 1 ) - 3}{2 ( - 2 )}

Remove parentheses:

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

= \frac{1 - 3}{- 2 \cdot 2}

Subtract the numbers:

1 - 3 = - 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}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = - 1, sin (θ) = \frac{1}{2}

sin (θ) = - 1, sin (θ) = \frac{1}{2}

sin (θ) = - 1, 0 \leq θ \leq 18 0^{\circ} :

No Solution

sin (θ) = - 1, 0 \leq θ \leq 18 0^{\circ}

General solutions for

sin (θ) = - 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 27 0^{\circ} + 36 0^{\circ} n

θ = 27 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq θ \leq 18 0^{\circ}

No Solution

sin (θ) = \frac{1}{2}, 0 \leq θ \leq 18 0^{\circ} : θ = 3 0^{\circ}, θ = 15 0^{\circ}

sin (θ) = \frac{1}{2}, 0 \leq θ \leq 18 0^{\circ}

General solutions for

sin (θ) = \frac{1}{2}

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq θ \leq 18 0^{\circ}

θ = 3 0^{\circ}, θ = 15 0^{\circ}

Combine all the solutions

θ = 3 0^{\circ}, θ = 15 0^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(θ)-1=sin(θ),0<= θ<= 180 ?
The general solution for 2cos^2(θ)-1=sin(θ),0<= θ<= 180 is θ=30,θ=150