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

The general solution for 4sin^2(θ)-sin(θ)-4=0 is θ=-1.08175…+2pin,θ=pi+1.08175…+2pin

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Popular >

4sin^2(θ)-sin(θ)-4=0

Solution

4 sin^{2} (θ) - sin (θ) - 4 = 0

Solution

θ = - 1.08175 \dots + 2 πn, θ = π + 1.08175 \dots + 2 πn

Degrees

θ = - 61.97982 \dots^{\circ} + 36 0^{\circ} n, θ = 241.97982 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 sin^{2} (θ) - sin (θ) - 4 = 0

Solve by substitution

4 sin^{2} (θ) - sin (θ) - 4 = 0

Let:

sin (θ) = u

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

4 u^{2} - u - 4 = 0 : u = \frac{1 + 65}{8}, u = \frac{1 - 65}{8}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = - 1, c = - 4

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

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

(- 1)^{2} - 4 \cdot 4 (- 4) = 65

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

Apply rule

- (- a) = a

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

(- 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 4 \cdot 4 = 64

4 \cdot 4 \cdot 4

Multiply the numbers:

4 \cdot 4 \cdot 4 = 64

= 64

= 1 + 64

Add the numbers:

1 + 64 = 65

= 65

u_{1, 2} = \frac{- ( - 1 ) \pm 65}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 65}{2 \cdot 4}, u_{2} = \frac{- ( - 1 ) - 65}{2 \cdot 4}

u = \frac{- ( - 1 ) + 65}{2 \cdot 4} : \frac{1 + 65}{8}

\frac{- ( - 1 ) + 65}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{1 + 65}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{1 + 65}{8}

u = \frac{- ( - 1 ) - 65}{2 \cdot 4} : \frac{1 - 65}{8}

\frac{- ( - 1 ) - 65}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{1 - 65}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{1 - 65}{8}

The solutions to the quadratic equation are:

u = \frac{1 + 65}{8}, u = \frac{1 - 65}{8}

Substitute back

u = sin (θ)

sin (θ) = \frac{1 + 65}{8}, sin (θ) = \frac{1 - 65}{8}

sin (θ) = \frac{1 + 65}{8}, sin (θ) = \frac{1 - 65}{8}

sin (θ) = \frac{1 + 65}{8} :

No Solution

sin (θ) = \frac{1 + 65}{8}

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = \frac{1 - 65}{8} : θ = arcsin (\frac{1 - 65}{8}) + 2 πn, θ = π + arcsin (- \frac{1 - 65}{8}) + 2 πn

sin (θ) = \frac{1 - 65}{8}

Apply trig inverse properties

sin (θ) = \frac{1 - 65}{8}

General solutions for

sin (θ) = \frac{1 - 65}{8}

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

θ = arcsin (\frac{1 - 65}{8}) + 2 πn, θ = π + arcsin (- \frac{1 - 65}{8}) + 2 πn

θ = arcsin (\frac{1 - 65}{8}) + 2 πn, θ = π + arcsin (- \frac{1 - 65}{8}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{1 - 65}{8}) + 2 πn, θ = π + arcsin (- \frac{1 - 65}{8}) + 2 πn

Show solutions in decimal form

θ = - 1.08175 \dots + 2 πn, θ = π + 1.08175 \dots + 2 πn

Popular Examples

tan(θ)=sqrt(1)

tan (θ) = 1

sin(2θ)+cos(2θ)=0

sin (2 θ) + cos (2 θ) = 0

5cot^2(x)-15=0

5 cot^{2} (x) - 15 = 0

sin(x)= 2/(sqrt(5))

sin (x) = \frac{2}{5}

csc(x)=5

csc (x) = 5

Frequently Asked Questions (FAQ)

What is the general solution for 4sin^2(θ)-sin(θ)-4=0 ?
The general solution for 4sin^2(θ)-sin(θ)-4=0 is θ=-1.08175…+2pin,θ=pi+1.08175…+2pin