What is the general solution for 5=20.39sin(θ)cos(θ) ?

The general solution for 5=20.39sin(θ)cos(θ) is θ=(0.51259…)/2+pin,θ= pi/2-(0.51259…)/2+pin

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Popular >

5=20.39sin(θ)cos(θ)

Solution

5 = 20.39 sin (θ) cos (θ)

Solution

θ = \frac{0.51259 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.51259 \dots}{2} + πn

Degrees

θ = 14.68463 \dots^{\circ} + 18 0^{\circ} n, θ = 75.31536 \dots^{\circ} + 18 0^{\circ} n

Solution steps

5 = 20.39 sin (θ) cos (θ)

Switch sides

20.39 sin (θ) cos (θ) = 5

Rewrite using trig identities

20.39 sin (θ) cos (θ)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

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

= 20.39 \cdot \frac{sin ( 2 θ )}{2}

20.39 \cdot \frac{sin ( 2 θ )}{2} = 5

Refine

20.39 \cdot \frac{sin ( 2 θ )}{2} : \frac{20.39 sin ( 2 θ )}{2}

20.39 \cdot \frac{sin ( 2 θ )}{2}

Multiply fractions:

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

= \frac{sin ( 2 θ ) \cdot 20.39}{2}

\frac{20.39 sin ( 2 θ )}{2} = 5

Multiply both sides by

2

\frac{20.39 sin ( 2 θ )}{2} = 5

Multiply both sides by

2

\frac{20.39 sin ( 2 θ )}{2} \cdot 2 = 5 \cdot 2

Simplify

\frac{20.39 sin ( 2 θ )}{2} \cdot 2 = 5 \cdot 2

Simplify

\frac{20.39 sin ( 2 θ )}{2} \cdot 2 : 20.39 sin (2 θ)

\frac{20.39 sin ( 2 θ )}{2} \cdot 2

Multiply fractions:

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

= \frac{20.39 sin ( 2 θ ) \cdot 2}{2}

Cancel the common factor:

2

= 20.39 sin (2 θ)

Simplify

5 \cdot 2 : 10

5 \cdot 2

Multiply the numbers:

5 \cdot 2 = 10

= 10

20.39 sin (2 θ) = 10

20.39 sin (2 θ) = 10

20.39 sin (2 θ) = 10

Divide both sides by

20.39

20.39 sin (2 θ) = 10

Divide both sides by

20.39

\frac{20.39 sin ( 2 θ )}{20.39} = \frac{10}{20.39}

Simplify

sin (2 θ) = 0.49043 \dots

sin (2 θ) = 0.49043 \dots

Apply trig inverse properties

sin (2 θ) = 0.49043 \dots

General solutions for

sin (2 θ) = 0.49043 \dots

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

2 θ = arcsin (0.49043 \dots) + 2 πn, 2 θ = π - arcsin (0.49043 \dots) + 2 πn

2 θ = arcsin (0.49043 \dots) + 2 πn, 2 θ = π - arcsin (0.49043 \dots) + 2 πn

Solve

2 θ = arcsin (0.49043 \dots) + 2 πn : θ = \frac{arcsin ( 0.49043 \dots )}{2} + πn

2 θ = arcsin (0.49043 \dots) + 2 πn

Divide both sides by

2

2 θ = arcsin (0.49043 \dots) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arcsin ( 0.49043 \dots )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{arcsin ( 0.49043 \dots )}{2} + πn

θ = \frac{arcsin ( 0.49043 \dots )}{2} + πn

Solve

2 θ = π - arcsin (0.49043 \dots) + 2 πn : θ = \frac{π}{2} - \frac{arcsin ( 0.49043 \dots )}{2} + πn

2 θ = π - arcsin (0.49043 \dots) + 2 πn

Divide both sides by

2

2 θ = π - arcsin (0.49043 \dots) + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{π}{2} - \frac{arcsin ( 0.49043 \dots )}{2} + \frac{2 πn}{2}

Simplify

θ = \frac{π}{2} - \frac{arcsin ( 0.49043 \dots )}{2} + πn

θ = \frac{π}{2} - \frac{arcsin ( 0.49043 \dots )}{2} + πn

θ = \frac{arcsin ( 0.49043 \dots )}{2} + πn, θ = \frac{π}{2} - \frac{arcsin ( 0.49043 \dots )}{2} + πn

Show solutions in decimal form

θ = \frac{0.51259 \dots}{2} + πn, θ = \frac{π}{2} - \frac{0.51259 \dots}{2} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5=20.39sin(θ)cos(θ) ?
The general solution for 5=20.39sin(θ)cos(θ) is θ=(0.51259…)/2+pin,θ= pi/2-(0.51259…)/2+pin