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

The general solution for 5sin(2θ)=-1/5 is θ=-(0.04001…)/2+pin,θ= pi/2+(0.04001…)/2+pin

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Popular >

5sin(2θ)=-1/5

Solution

5 sin (2 θ) = - \frac{1}{5}

Solution

θ = - \frac{0.04001 \dots}{2} + πn, θ = \frac{π}{2} + \frac{0.04001 \dots}{2} + πn

Degrees

θ = - 1.14622 \dots^{\circ} + 18 0^{\circ} n, θ = 91.14622 \dots^{\circ} + 18 0^{\circ} n

Solution steps

5 sin (2 θ) = - \frac{1}{5}

Divide both sides by

5

5 sin (2 θ) = - \frac{1}{5}

Divide both sides by

5

\frac{5 sin ( 2 θ )}{5} = \frac{- \frac{1}{5}}{5}

Simplify

\frac{5 sin ( 2 θ )}{5} = \frac{- \frac{1}{5}}{5}

Simplify

\frac{5 sin ( 2 θ )}{5} : sin (2 θ)

\frac{5 sin ( 2 θ )}{5}

Divide the numbers:

\frac{5}{5} = 1

= sin (2 θ)

Simplify

\frac{- \frac{1}{5}}{5} : - \frac{1}{25}

\frac{- \frac{1}{5}}{5}

Apply the fraction rule:

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

= - \frac{\frac{1}{5}}{5}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

\frac{\frac{1}{5}}{5} = \frac{1}{5 \cdot 5}

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

Multiply the numbers:

5 \cdot 5 = 25

= - \frac{1}{25}

sin (2 θ) = - \frac{1}{25}

sin (2 θ) = - \frac{1}{25}

sin (2 θ) = - \frac{1}{25}

Apply trig inverse properties

sin (2 θ) = - \frac{1}{25}

General solutions for

sin (2 θ) = - \frac{1}{25}

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

2 θ = arcsin (- \frac{1}{25}) + 2 πn, 2 θ = π + arcsin (\frac{1}{25}) + 2 πn

2 θ = arcsin (- \frac{1}{25}) + 2 πn, 2 θ = π + arcsin (\frac{1}{25}) + 2 πn

Solve

2 θ = arcsin (- \frac{1}{25}) + 2 πn : θ = - \frac{arcsin ( \frac{1}{25} )}{2} + πn

2 θ = arcsin (- \frac{1}{25}) + 2 πn

Simplify

arcsin (- \frac{1}{25}) + 2 πn : - arcsin (\frac{1}{25}) + 2 πn

arcsin (- \frac{1}{25}) + 2 πn

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{25}) = - arcsin (\frac{1}{25})

= - arcsin (\frac{1}{25}) + 2 πn

2 θ = - arcsin (\frac{1}{25}) + 2 πn

Divide both sides by

2

2 θ = - arcsin (\frac{1}{25}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = - \frac{arcsin ( \frac{1}{25} )}{2} + \frac{2 πn}{2}

Simplify

θ = - \frac{arcsin ( \frac{1}{25} )}{2} + πn

θ = - \frac{arcsin ( \frac{1}{25} )}{2} + πn

Solve

2 θ = π + arcsin (\frac{1}{25}) + 2 πn : θ = \frac{π}{2} + \frac{arcsin ( \frac{1}{25} )}{2} + πn

2 θ = π + arcsin (\frac{1}{25}) + 2 πn

Divide both sides by

2

2 θ = π + arcsin (\frac{1}{25}) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{π}{2} + \frac{arcsin ( \frac{1}{25} )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{π}{2} + \frac{arcsin ( \frac{1}{25} )}{2} + πn

θ = \frac{π}{2} + \frac{arcsin ( \frac{1}{25} )}{2} + πn

θ = - \frac{arcsin ( \frac{1}{25} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( \frac{1}{25} )}{2} + πn

Show solutions in decimal form

θ = - \frac{0.04001 \dots}{2} + πn, θ = \frac{π}{2} + \frac{0.04001 \dots}{2} + πn

Popular Examples

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sin (θ) = \frac{8}{17}, cos (θ) = \frac{15}{17}, tan (θ)

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

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

sin(θ)=-12/13 ,sec(θ)

sin (θ) = - \frac{12}{13}, sec (θ)

tan(x)= 7/10

tan (x) = \frac{7}{10}

3sec(x)=3

3 sec (x) = 3

Frequently Asked Questions (FAQ)

What is the general solution for 5sin(2θ)=-1/5 ?
The general solution for 5sin(2θ)=-1/5 is θ=-(0.04001…)/2+pin,θ= pi/2+(0.04001…)/2+pin