What is the general solution for -4sin^2(θ)-7sin(θ)+7=0 ?

The general solution for -4sin^2(θ)-7sin(θ)+7=0 is θ=0.79102…+2pin,θ=pi-0.79102…+2pin

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Popular >

-4sin^2(θ)-7sin(θ)+7=0

Solution

- 4 sin^{2} (θ) - 7 sin (θ) + 7 = 0

Solution

θ = 0.79102 \dots + 2 πn, θ = π - 0.79102 \dots + 2 πn

Degrees

θ = 45.32221 \dots^{\circ} + 36 0^{\circ} n, θ = 134.67778 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 4 sin^{2} (θ) - 7 sin (θ) + 7 = 0

Solve by substitution

- 4 sin^{2} (θ) - 7 sin (θ) + 7 = 0

Let:

sin (θ) = u

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

- 4 u^{2} - 7 u + 7 = 0 : u = - \frac{7 + 161}{8}, u = \frac{161 - 7}{8}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 4, b = - 7, c = 7

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

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

(- 7)^{2} - 4 (- 4) \cdot 7 = 161

(- 7)^{2} - 4 (- 4) \cdot 7

Apply rule

- (- a) = a

= (- 7)^{2} + 4 \cdot 4 \cdot 7

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} + 4 \cdot 4 \cdot 7

Multiply the numbers:

4 \cdot 4 \cdot 7 = 112

= 7^{2} + 112

7^{2} = 49

= 49 + 112

Add the numbers:

49 + 112 = 161

= 161

u_{1, 2} = \frac{- ( - 7 ) \pm 161}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 161}{2 ( - 4 )}, u_{2} = \frac{- ( - 7 ) - 161}{2 ( - 4 )}

u = \frac{- ( - 7 ) + 161}{2 ( - 4 )} : - \frac{7 + 161}{8}

\frac{- ( - 7 ) + 161}{2 ( - 4 )}

Remove parentheses:

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

= \frac{7 + 161}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{7 + 161}{- 8}

Apply the fraction rule:

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

= - \frac{7 + 161}{8}

u = \frac{- ( - 7 ) - 161}{2 ( - 4 )} : \frac{161 - 7}{8}

\frac{- ( - 7 ) - 161}{2 ( - 4 )}

Remove parentheses:

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

= \frac{7 - 161}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{7 - 161}{- 8}

Apply the fraction rule:

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

7 - 161 = - (161 - 7)

= \frac{161 - 7}{8}

The solutions to the quadratic equation are:

u = - \frac{7 + 161}{8}, u = \frac{161 - 7}{8}

Substitute back

u = sin (θ)

sin (θ) = - \frac{7 + 161}{8}, sin (θ) = \frac{161 - 7}{8}

sin (θ) = - \frac{7 + 161}{8}, sin (θ) = \frac{161 - 7}{8}

sin (θ) = - \frac{7 + 161}{8} :

No Solution

sin (θ) = - \frac{7 + 161}{8}

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = \frac{161 - 7}{8} : θ = arcsin (\frac{161 - 7}{8}) + 2 πn, θ = π - arcsin (\frac{161 - 7}{8}) + 2 πn

sin (θ) = \frac{161 - 7}{8}

Apply trig inverse properties

sin (θ) = \frac{161 - 7}{8}

General solutions for

sin (θ) = \frac{161 - 7}{8}

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

θ = arcsin (\frac{161 - 7}{8}) + 2 πn, θ = π - arcsin (\frac{161 - 7}{8}) + 2 πn

θ = arcsin (\frac{161 - 7}{8}) + 2 πn, θ = π - arcsin (\frac{161 - 7}{8}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{161 - 7}{8}) + 2 πn, θ = π - arcsin (\frac{161 - 7}{8}) + 2 πn

Show solutions in decimal form

θ = 0.79102 \dots + 2 πn, θ = π - 0.79102 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -4sin^2(θ)-7sin(θ)+7=0 ?
The general solution for -4sin^2(θ)-7sin(θ)+7=0 is θ=0.79102…+2pin,θ=pi-0.79102…+2pin