What is the general solution for -4sin^2(θ)-3sin(θ)+5=0 ?

The general solution for -4sin^2(θ)-3sin(θ)+5=0 is θ=0.93440…+2pin,θ=pi-0.93440…+2pin

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Popular >

-4sin^2(θ)-3sin(θ)+5=0

Solution

- 4 sin^{2} (θ) - 3 sin (θ) + 5 = 0

Solution

θ = 0.93440 \dots + 2 πn, θ = π - 0.93440 \dots + 2 πn

Degrees

θ = 53.53765 \dots^{\circ} + 36 0^{\circ} n, θ = 126.46234 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 4 sin^{2} (θ) - 3 sin (θ) + 5 = 0

Solve by substitution

- 4 sin^{2} (θ) - 3 sin (θ) + 5 = 0

Let:

sin (θ) = u

- 4 u^{2} - 3 u + 5 = 0

- 4 u^{2} - 3 u + 5 = 0 : u = - \frac{3 + 89}{8}, u = \frac{89 - 3}{8}

- 4 u^{2} - 3 u + 5 = 0

Solve with the quadratic formula

- 4 u^{2} - 3 u + 5 = 0

Quadratic Equation Formula:

For

a = - 4, b = - 3, c = 5

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( - 4 ) \cdot 5}{2 ( - 4 )}

(- 3)^{2} - 4 (- 4) \cdot 5 = 89

(- 3)^{2} - 4 (- 4) \cdot 5

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 4 \cdot 5

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 4 \cdot 5

Multiply the numbers:

4 \cdot 4 \cdot 5 = 80

= 3^{2} + 80

3^{2} = 9

= 9 + 80

Add the numbers:

9 + 80 = 89

= 89

u_{1, 2} = \frac{- ( - 3 ) \pm 89}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 89}{2 ( - 4 )}, u_{2} = \frac{- ( - 3 ) - 89}{2 ( - 4 )}

u = \frac{- ( - 3 ) + 89}{2 ( - 4 )} : - \frac{3 + 89}{8}

\frac{- ( - 3 ) + 89}{2 ( - 4 )}

Remove parentheses:

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

= \frac{3 + 89}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{3 + 89}{- 8}

Apply the fraction rule:

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

= - \frac{3 + 89}{8}

u = \frac{- ( - 3 ) - 89}{2 ( - 4 )} : \frac{89 - 3}{8}

\frac{- ( - 3 ) - 89}{2 ( - 4 )}

Remove parentheses:

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

= \frac{3 - 89}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{3 - 89}{- 8}

Apply the fraction rule:

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

3 - 89 = - (89 - 3)

= \frac{89 - 3}{8}

The solutions to the quadratic equation are:

u = - \frac{3 + 89}{8}, u = \frac{89 - 3}{8}

Substitute back

u = sin (θ)

sin (θ) = - \frac{3 + 89}{8}, sin (θ) = \frac{89 - 3}{8}

sin (θ) = - \frac{3 + 89}{8}, sin (θ) = \frac{89 - 3}{8}

sin (θ) = - \frac{3 + 89}{8} :

No Solution

sin (θ) = - \frac{3 + 89}{8}

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = \frac{89 - 3}{8} : θ = arcsin (\frac{89 - 3}{8}) + 2 πn, θ = π - arcsin (\frac{89 - 3}{8}) + 2 πn

sin (θ) = \frac{89 - 3}{8}

Apply trig inverse properties

sin (θ) = \frac{89 - 3}{8}

General solutions for

sin (θ) = \frac{89 - 3}{8}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{89 - 3}{8}) + 2 πn, θ = π - arcsin (\frac{89 - 3}{8}) + 2 πn

θ = arcsin (\frac{89 - 3}{8}) + 2 πn, θ = π - arcsin (\frac{89 - 3}{8}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{89 - 3}{8}) + 2 πn, θ = π - arcsin (\frac{89 - 3}{8}) + 2 πn

Show solutions in decimal form

θ = 0.93440 \dots + 2 πn, θ = π - 0.93440 \dots + 2 πn

Popular Examples

cos^2(x)-sin^2(x)=-1/2

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

cos(θ)+2cos(θ+240)=1

cos (θ) + 2 cos (θ + 24 0^{\circ}) = 1

arccos(x)= pi/8

arccos (x) = \frac{π}{8}

sin(θ)=5

sin (θ) = 5

sin^2(x)=4cos(x)

sin^{2} (x) = 4 cos (x)

Frequently Asked Questions (FAQ)

What is the general solution for -4sin^2(θ)-3sin(θ)+5=0 ?
The general solution for -4sin^2(θ)-3sin(θ)+5=0 is θ=0.93440…+2pin,θ=pi-0.93440…+2pin