What is the general solution for 5sin(2θ)-7sin(θ)=0 ?

The general solution for 5sin(2θ)-7sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=0.79539…+2pin,θ=2pi-0.79539…+2pin

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Popular >

5sin(2θ)-7sin(θ)=0

Solution

5 sin (2 θ) - 7 sin (θ) = 0

Solution

θ = 2 πn, θ = π + 2 πn, θ = 0.79539 \dots + 2 πn, θ = 2 π - 0.79539 \dots + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 45.57299 \dots^{\circ} + 36 0^{\circ} n, θ = 314.42700 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin (2 θ) - 7 sin (θ) = 0

Rewrite using trig identities

5 sin (2 θ) - 7 sin (θ)

Use the Double Angle identity:

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

= 5 \cdot 2 sin (θ) cos (θ) - 7 sin (θ)

Simplify

= 10 sin (θ) cos (θ) - 7 sin (θ)

- 7 sin (θ) + 10 cos (θ) sin (θ) = 0

Factor

- 7 sin (θ) + 10 cos (θ) sin (θ) : sin (θ) (10 cos (θ) - 7)

- 7 sin (θ) + 10 cos (θ) sin (θ)

Factor out common term

sin (θ)

= sin (θ) (- 7 + 10 cos (θ))

sin (θ) (10 cos (θ) - 7) = 0

Solving each part separately

sin (θ) = 0 or 10 cos (θ) - 7 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

10 cos (θ) - 7 = 0 : θ = arccos (\frac{7}{10}) + 2 πn, θ = 2 π - arccos (\frac{7}{10}) + 2 πn

10 cos (θ) - 7 = 0

Move

7

to the right side

10 cos (θ) - 7 = 0

Add

7

to both sides

10 cos (θ) - 7 + 7 = 0 + 7

Simplify

10 cos (θ) = 7

10 cos (θ) = 7

Divide both sides by

10

10 cos (θ) = 7

Divide both sides by

10

\frac{10 cos ( θ )}{10} = \frac{7}{10}

Simplify

cos (θ) = \frac{7}{10}

cos (θ) = \frac{7}{10}

Apply trig inverse properties

cos (θ) = \frac{7}{10}

General solutions for

cos (θ) = \frac{7}{10}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{7}{10}) + 2 πn, θ = 2 π - arccos (\frac{7}{10}) + 2 πn

θ = arccos (\frac{7}{10}) + 2 πn, θ = 2 π - arccos (\frac{7}{10}) + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arccos (\frac{7}{10}) + 2 πn, θ = 2 π - arccos (\frac{7}{10}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = 0.79539 \dots + 2 πn, θ = 2 π - 0.79539 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 5sin(2θ)-7sin(θ)=0 ?
The general solution for 5sin(2θ)-7sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=0.79539…+2pin,θ=2pi-0.79539…+2pin