What is the general solution for -4sin^2(θ)-15sin(θ)=-7sin(θ)-5 ?

The general solution for -4sin^2(θ)-15sin(θ)=-7sin(θ)-5 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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Popular >

-4sin^2(θ)-15sin(θ)=-7sin(θ)-5

Solution

- 4 sin^{2} (θ) - 15 sin (θ) = - 7 sin (θ) - 5

Solution

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Degrees

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

Solution steps

- 4 sin^{2} (θ) - 15 sin (θ) = - 7 sin (θ) - 5

Solve by substitution

- 4 sin^{2} (θ) - 15 sin (θ) = - 7 sin (θ) - 5

Let:

sin (θ) = u

- 4 u^{2} - 15 u = - 7 u - 5

- 4 u^{2} - 15 u = - 7 u - 5 : u = - \frac{5}{2}, u = \frac{1}{2}

- 4 u^{2} - 15 u = - 7 u - 5

Move

5

to the left side

- 4 u^{2} - 15 u = - 7 u - 5

Add

5

to both sides

- 4 u^{2} - 15 u + 5 = - 7 u - 5 + 5

Simplify

- 4 u^{2} - 15 u + 5 = - 7 u

- 4 u^{2} - 15 u + 5 = - 7 u

Move

7 u

to the left side

- 4 u^{2} - 15 u + 5 = - 7 u

Add

7 u

to both sides

- 4 u^{2} - 15 u + 5 + 7 u = - 7 u + 7 u

Simplify

- 4 u^{2} - 8 u + 5 = 0

- 4 u^{2} - 8 u + 5 = 0

Solve with the quadratic formula

- 4 u^{2} - 8 u + 5 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 8, c = 5

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

(- 8)^{2} - 4 (- 4) \cdot 5 = 12

(- 8)^{2} - 4 (- 4) \cdot 5

Apply rule

- (- a) = a

= (- 8)^{2} + 4 \cdot 4 \cdot 5

Apply exponent rule:

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

n

is even

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 4 \cdot 5

Multiply the numbers:

4 \cdot 4 \cdot 5 = 80

= 8^{2} + 80

8^{2} = 64

= 64 + 80

Add the numbers:

64 + 80 = 144

= 144

Factor the number:

144 = 1 2^{2}

= 1 2^{2}

Apply radical rule:

1 2^{2} = 12

= 12

u_{1, 2} = \frac{- ( - 8 ) \pm 12}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 8 ) + 12}{2 ( - 4 )}, u_{2} = \frac{- ( - 8 ) - 12}{2 ( - 4 )}

u = \frac{- ( - 8 ) + 12}{2 ( - 4 )} : - \frac{5}{2}

\frac{- ( - 8 ) + 12}{2 ( - 4 )}

Remove parentheses:

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

= \frac{8 + 12}{- 2 \cdot 4}

Add the numbers:

8 + 12 = 20

= \frac{20}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{20}{- 8}

Apply the fraction rule:

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

= - \frac{20}{8}

Cancel the common factor:

4

= - \frac{5}{2}

u = \frac{- ( - 8 ) - 12}{2 ( - 4 )} : \frac{1}{2}

\frac{- ( - 8 ) - 12}{2 ( - 4 )}

Remove parentheses:

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

= \frac{8 - 12}{- 2 \cdot 4}

Subtract the numbers:

8 - 12 = - 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

The solutions to the quadratic equation are:

u = - \frac{5}{2}, u = \frac{1}{2}

Substitute back

u = sin (θ)

sin (θ) = - \frac{5}{2}, sin (θ) = \frac{1}{2}

sin (θ) = - \frac{5}{2}, sin (θ) = \frac{1}{2}

sin (θ) = - \frac{5}{2} :

No Solution

sin (θ) = - \frac{5}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 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{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Popular Examples

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tan (θ) = \frac{2 3}{1}

cos^2(x)-cos(x)=0,x,0<= x<= 2pi

cos^{2} (x) - cos (x) = 0, x, 0 \leq x \leq 2 π

6cos^2(x)+sin(x)-4=0

6 cos^{2} (x) + sin (x) - 4 = 0

2sin(4x)-sqrt(3)=0

2 sin (4 x) - 3 = 0

Frequently Asked Questions (FAQ)

What is the general solution for -4sin^2(θ)-15sin(θ)=-7sin(θ)-5 ?
The general solution for -4sin^2(θ)-15sin(θ)=-7sin(θ)-5 is θ= pi/6+2pin,θ=(5pi)/6+2pin