What is the general solution for -2sin^2(θ)+10sin(θ)-3=3sin(θ),0<= θ<360 ?

Q: What is the general solution for -2sin^2(θ)+10sin(θ)-3=3sin(θ),0<= θ<360 ?

The general solution for -2sin^2(θ)+10sin(θ)-3=3sin(θ),0<= θ<360 is θ=30,θ=150

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-2sin^2(θ)+10sin(θ)-3=3sin(θ),0<= θ<360

Solution

- 2 sin^{2} (θ) + 10 sin (θ) - 3 = 3 sin (θ), 0^{\circ} \leq θ < 36 0^{\circ}

Solution

θ = 3 0^{\circ}, θ = 15 0^{\circ}

Radians

θ = \frac{π}{6}, θ = \frac{5 π}{6}

Solution steps

- 2 sin^{2} (θ) + 10 sin (θ) - 3 = 3 sin (θ), 0^{\circ} \leq θ < 36 0^{\circ}

Solve by substitution

- 2 sin^{2} (θ) + 10 sin (θ) - 3 = 3 sin (θ)

Let:

sin (θ) = u

- 2 u^{2} + 10 u - 3 = 3 u

- 2 u^{2} + 10 u - 3 = 3 u : u = \frac{1}{2}, u = 3

- 2 u^{2} + 10 u - 3 = 3 u

Move

3 u

to the left side

- 2 u^{2} + 10 u - 3 = 3 u

Subtract

3 u

from both sides

- 2 u^{2} + 10 u - 3 - 3 u = 3 u - 3 u

Simplify

- 2 u^{2} + 7 u - 3 = 0

- 2 u^{2} + 7 u - 3 = 0

Solve with the quadratic formula

- 2 u^{2} + 7 u - 3 = 0

Quadratic Equation Formula:

For

a = - 2, b = 7, c = - 3

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 2 ) ( - 3 )}{2 ( - 2 )}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 2 ) ( - 3 )}{2 ( - 2 )}

7^{2} - 4 (- 2) (- 3) = 5

7^{2} - 4 (- 2) (- 3)

Apply rule

- (- a) = a

= 7^{2} - 4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 7^{2} - 24

7^{2} = 49

= 49 - 24

Subtract the numbers:

49 - 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 7 \pm 5}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 7 + 5}{2 ( - 2 )}, u_{2} = \frac{- 7 - 5}{2 ( - 2 )}

u = \frac{- 7 + 5}{2 ( - 2 )} : \frac{1}{2}

\frac{- 7 + 5}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 7 + 5}{- 2 \cdot 2}

Add/Subtract the numbers:

- 7 + 5 = - 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{- 7 - 5}{2 ( - 2 )} : 3

\frac{- 7 - 5}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 7 - 5}{- 2 \cdot 2}

Subtract the numbers:

- 7 - 5 = - 12

= \frac{- 12}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 12}{- 4}

Apply the fraction rule:

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

= \frac{12}{4}

Divide the numbers:

\frac{12}{4} = 3

= 3

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = 3

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = 3

sin (θ) = \frac{1}{2}, sin (θ) = 3

sin (θ) = \frac{1}{2}, 0 \leq θ < 36 0^{\circ} : θ = 3 0^{\circ}, θ = 15 0^{\circ}

sin (θ) = \frac{1}{2}, 0 \leq θ < 36 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 < 36 0^{\circ}

θ = 3 0^{\circ}, θ = 15 0^{\circ}

sin (θ) = 3, 0 \leq θ < 36 0^{\circ} :

No Solution

sin (θ) = 3, 0 \leq θ < 36 0^{\circ}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

θ = 3 0^{\circ}, θ = 15 0^{\circ}

Popular Examples

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cot^{2} (θ) + 4 csc (θ) = - 5

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10 sin (a) + 7 sin (a) + 1 sin (a) + 8 sin (a) = 48

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2 cos^{2} (θ) + 7 sin (θ) = 5

(2sqrt(3))/3 =sec(θ)

\frac{2 3}{3} = sec (θ)

4(sin(x))^2-15sin(x)-4=0

4 (sin (x))^{2} - 15 sin (x) - 4 = 0

Frequently Asked Questions (FAQ)

What is the general solution for -2sin^2(θ)+10sin(θ)-3=3sin(θ),0<= θ<360 ?
The general solution for -2sin^2(θ)+10sin(θ)-3=3sin(θ),0<= θ<360 is θ=30,θ=150