What is the general solution for 7sin(θ)=7cos(2θ) ?

The general solution for 7sin(θ)=7cos(2θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin

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7sin(θ)=7cos(2θ)

Solution

7 sin (θ) = 7 cos (2 θ)

Solution

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

7 sin (θ) = 7 cos (2 θ)

Subtract

7 cos (2 θ)

from both sides

7 sin (θ) - 7 cos (2 θ) = 0

Rewrite using trig identities

- 7 cos (2 θ) + 7 sin (θ)

Use the Double Angle identity:

cos (2 x) = 1 - 2 sin^{2} (x)

= - 7 (1 - 2 sin^{2} (θ)) + 7 sin (θ)

- (1 - 2 sin^{2} (θ)) \cdot 7 + 7 sin (θ) = 0

Solve by substitution

- (1 - 2 sin^{2} (θ)) \cdot 7 + 7 sin (θ) = 0

Let:

sin (θ) = u

- (1 - 2 u^{2}) \cdot 7 + 7 u = 0

- (1 - 2 u^{2}) \cdot 7 + 7 u = 0 : u = \frac{1}{2}, u = - 1

- (1 - 2 u^{2}) \cdot 7 + 7 u = 0

Expand

- (1 - 2 u^{2}) \cdot 7 + 7 u : - 7 + 14 u^{2} + 7 u

- (1 - 2 u^{2}) \cdot 7 + 7 u

= - 7 (1 - 2 u^{2}) + 7 u

Expand

- 7 (1 - 2 u^{2}) : - 7 + 14 u^{2}

- 7 (1 - 2 u^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = - 7, b = 1, c = 2 u^{2}

= - 7 \cdot 1 - (- 7) \cdot 2 u^{2}

Apply minus-plus rules

- (- a) = a

= - 7 \cdot 1 + 7 \cdot 2 u^{2}

Simplify

- 7 \cdot 1 + 7 \cdot 2 u^{2} : - 7 + 14 u^{2}

- 7 \cdot 1 + 7 \cdot 2 u^{2}

Multiply the numbers:

7 \cdot 1 = 7

= - 7 + 7 \cdot 2 u^{2}

Multiply the numbers:

7 \cdot 2 = 14

= - 7 + 14 u^{2}

= - 7 + 14 u^{2}

= - 7 + 14 u^{2} + 7 u

- 7 + 14 u^{2} + 7 u = 0

Write in the standard form

a x^{2} + b x + c = 0

14 u^{2} + 7 u - 7 = 0

Solve with the quadratic formula

14 u^{2} + 7 u - 7 = 0

Quadratic Equation Formula:

For

a = 14, b = 7, c = - 7

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 14 ( - 7 )}{2 \cdot 14}

u_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 14 ( - 7 )}{2 \cdot 14}

7^{2} - 4 \cdot 14 (- 7) = 21

7^{2} - 4 \cdot 14 (- 7)

Apply rule

- (- a) = a

= 7^{2} + 4 \cdot 14 \cdot 7

Multiply the numbers:

4 \cdot 14 \cdot 7 = 392

= 7^{2} + 392

7^{2} = 49

= 49 + 392

Add the numbers:

49 + 392 = 441

= 441

Factor the number:

441 = 2 1^{2}

= 2 1^{2}

Apply radical rule:

n a^{n} = a

2 1^{2} = 21

= 21

u_{1, 2} = \frac{- 7 \pm 21}{2 \cdot 14}

Separate the solutions

u_{1} = \frac{- 7 + 21}{2 \cdot 14}, u_{2} = \frac{- 7 - 21}{2 \cdot 14}

u = \frac{- 7 + 21}{2 \cdot 14} : \frac{1}{2}

\frac{- 7 + 21}{2 \cdot 14}

Add/Subtract the numbers:

- 7 + 21 = 14

= \frac{14}{2 \cdot 14}

Multiply the numbers:

2 \cdot 14 = 28

= \frac{14}{28}

Cancel the common factor:

14

= \frac{1}{2}

u = \frac{- 7 - 21}{2 \cdot 14} : - 1

\frac{- 7 - 21}{2 \cdot 14}

Subtract the numbers:

- 7 - 21 = - 28

= \frac{- 28}{2 \cdot 14}

Multiply the numbers:

2 \cdot 14 = 28

= \frac{- 28}{28}

Apply the fraction rule:

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

= - \frac{28}{28}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = - 1

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = - 1

sin (θ) = \frac{1}{2}, sin (θ) = - 1

sin (θ) = \frac{1}{2} : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

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 = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

sin (θ) = - 1

General solutions for

sin (θ) = - 1

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 = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 7sin(θ)=7cos(2θ) ?
The general solution for 7sin(θ)=7cos(2θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin