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

The general solution for 2sin^2(θ)-sin(θ)-2=0 is θ=-0.89590…+2pin,θ=pi+0.89590…+2pin

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Popular >

2sin^2(θ)-sin(θ)-2=0

Solution

2 sin^{2} (θ) - sin (θ) - 2 = 0

Solution

θ = - 0.89590 \dots + 2 πn, θ = π + 0.89590 \dots + 2 πn

Degrees

θ = - 51.33171 \dots^{\circ} + 36 0^{\circ} n, θ = 231.33171 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (θ) - sin (θ) - 2 = 0

Solve by substitution

2 sin^{2} (θ) - sin (θ) - 2 = 0

Let:

sin (θ) = u

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

2 u^{2} - u - 2 = 0 : u = \frac{1 + 17}{4}, u = \frac{1 - 17}{4}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 2, b = - 1, c = - 2

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

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

(- 1)^{2} - 4 \cdot 2 (- 2) = 17

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

Apply rule

- (- a) = a

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

(- 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 2 \cdot 2 = 16

4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 16

= 1 + 16

Add the numbers:

1 + 16 = 17

= 17

u_{1, 2} = \frac{- ( - 1 ) \pm 17}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 17}{2 \cdot 2}, u_{2} = \frac{- ( - 1 ) - 17}{2 \cdot 2}

u = \frac{- ( - 1 ) + 17}{2 \cdot 2} : \frac{1 + 17}{4}

\frac{- ( - 1 ) + 17}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 17}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 + 17}{4}

u = \frac{- ( - 1 ) - 17}{2 \cdot 2} : \frac{1 - 17}{4}

\frac{- ( - 1 ) - 17}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 17}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{1 - 17}{4}

The solutions to the quadratic equation are:

u = \frac{1 + 17}{4}, u = \frac{1 - 17}{4}

Substitute back

u = sin (θ)

sin (θ) = \frac{1 + 17}{4}, sin (θ) = \frac{1 - 17}{4}

sin (θ) = \frac{1 + 17}{4}, sin (θ) = \frac{1 - 17}{4}

sin (θ) = \frac{1 + 17}{4} :

No Solution

sin (θ) = \frac{1 + 17}{4}

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = \frac{1 - 17}{4} : θ = arcsin (\frac{1 - 17}{4}) + 2 πn, θ = π + arcsin (- \frac{1 - 17}{4}) + 2 πn

sin (θ) = \frac{1 - 17}{4}

Apply trig inverse properties

sin (θ) = \frac{1 - 17}{4}

General solutions for

sin (θ) = \frac{1 - 17}{4}

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

θ = arcsin (\frac{1 - 17}{4}) + 2 πn, θ = π + arcsin (- \frac{1 - 17}{4}) + 2 πn

θ = arcsin (\frac{1 - 17}{4}) + 2 πn, θ = π + arcsin (- \frac{1 - 17}{4}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{1 - 17}{4}) + 2 πn, θ = π + arcsin (- \frac{1 - 17}{4}) + 2 πn

Show solutions in decimal form

θ = - 0.89590 \dots + 2 πn, θ = π + 0.89590 \dots + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(θ)-sin(θ)-2=0 ?
The general solution for 2sin^2(θ)-sin(θ)-2=0 is θ=-0.89590…+2pin,θ=pi+0.89590…+2pin