What is the general solution for sin^2(2θ)-5sin(2θ)-1=0 ?

The general solution for sin^2(2θ)-5sin(2θ)-1=0 is θ=(-0.19379…)/2+pin,θ= pi/2+(0.19379…)/2+pin

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Popular >

sin^2(2θ)-5sin(2θ)-1=0

Solution

sin^{2} (2 θ) - 5 sin (2 θ) - 1 = 0

Solution

θ = \frac{- 0.19379 \dots}{2} + πn, θ = \frac{π}{2} + \frac{0.19379 \dots}{2} + πn

Degrees

θ = - 5.55176 \dots^{\circ} + 18 0^{\circ} n, θ = 95.55176 \dots^{\circ} + 18 0^{\circ} n

Solution steps

sin^{2} (2 θ) - 5 sin (2 θ) - 1 = 0

Solve by substitution

sin^{2} (2 θ) - 5 sin (2 θ) - 1 = 0

Let:

sin (2 θ) = u

u^{2} - 5 u - 1 = 0

u^{2} - 5 u - 1 = 0 : u = \frac{5 + 29}{2}, u = \frac{5 - 29}{2}

u^{2} - 5 u - 1 = 0

Solve with the quadratic formula

u^{2} - 5 u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = - 5, c = - 1

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

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

(- 5)^{2} - 4 \cdot 1 \cdot (- 1) = 29

(- 5)^{2} - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= (- 5)^{2} + 4 \cdot 1 \cdot 1

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 5^{2} + 4

5^{2} = 25

= 25 + 4

Add the numbers:

25 + 4 = 29

= 29

u_{1, 2} = \frac{- ( - 5 ) \pm 29}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 29}{2 \cdot 1}, u_{2} = \frac{- ( - 5 ) - 29}{2 \cdot 1}

u = \frac{- ( - 5 ) + 29}{2 \cdot 1} : \frac{5 + 29}{2}

\frac{- ( - 5 ) + 29}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{5 + 29}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{5 + 29}{2}

u = \frac{- ( - 5 ) - 29}{2 \cdot 1} : \frac{5 - 29}{2}

\frac{- ( - 5 ) - 29}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{5 - 29}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{5 - 29}{2}

The solutions to the quadratic equation are:

u = \frac{5 + 29}{2}, u = \frac{5 - 29}{2}

Substitute back

u = sin (2 θ)

sin (2 θ) = \frac{5 + 29}{2}, sin (2 θ) = \frac{5 - 29}{2}

sin (2 θ) = \frac{5 + 29}{2}, sin (2 θ) = \frac{5 - 29}{2}

sin (2 θ) = \frac{5 + 29}{2} :

No Solution

sin (2 θ) = \frac{5 + 29}{2}

- 1 \leq sin (x) \leq 1

No Solution

sin (2 θ) = \frac{5 - 29}{2} : θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

sin (2 θ) = \frac{5 - 29}{2}

Apply trig inverse properties

sin (2 θ) = \frac{5 - 29}{2}

General solutions for

sin (2 θ) = \frac{5 - 29}{2}

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

2 θ = arcsin (\frac{5 - 29}{2}) + 2 πn, 2 θ = π + arcsin (- \frac{5 - 29}{2}) + 2 πn

2 θ = arcsin (\frac{5 - 29}{2}) + 2 πn, 2 θ = π + arcsin (- \frac{5 - 29}{2}) + 2 πn

Solve

2 θ = arcsin (\frac{5 - 29}{2}) + 2 πn : θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn

2 θ = arcsin (\frac{5 - 29}{2}) + 2 πn

Divide both sides by

2

2 θ = arcsin (\frac{5 - 29}{2}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn

θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn

Solve

2 θ = π + arcsin (- \frac{5 - 29}{2}) + 2 πn : θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

2 θ = π + arcsin (- \frac{5 - 29}{2}) + 2 πn

- \frac{5 - 29}{2} = - \frac{- ( 29 - 5 )}{2} = \frac{29 - 5}{2}

2 θ = π + arcsin (\frac{29 - 5}{2}) + 2 πn

Divide both sides by

2

2 θ = π + arcsin (\frac{29 - 5}{2}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

Combine all the solutions

θ = \frac{arcsin ( \frac{5 - 29}{2} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( \frac{29 - 5}{2} )}{2} + πn

Show solutions in decimal form

θ = \frac{- 0.19379 \dots}{2} + πn, θ = \frac{π}{2} + \frac{0.19379 \dots}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(2θ)-5sin(2θ)-1=0 ?
The general solution for sin^2(2θ)-5sin(2θ)-1=0 is θ=(-0.19379…)/2+pin,θ= pi/2+(0.19379…)/2+pin