What is the general solution for 3sin(ϕ)-cos(2ϕ)=1 ?

The general solution for 3sin(ϕ)-cos(2ϕ)=1 is ϕ= pi/6+2pin,ϕ=(5pi)/6+2pin

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3sin(ϕ)-cos(2ϕ)=1

Solution

3 sin (ϕ) - cos (2 ϕ) = 1

Solution

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

Degrees

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

Solution steps

3 sin (ϕ) - cos (2 ϕ) = 1

Subtract

1

from both sides

3 sin (ϕ) - cos (2 ϕ) - 1 = 0

Rewrite using trig identities

- 1 - cos (2 ϕ) + 3 sin (ϕ)

Use the Double Angle identity:

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

= - 1 - (1 - 2 sin^{2} (ϕ)) + 3 sin (ϕ)

Simplify

- 1 - (1 - 2 sin^{2} (ϕ)) + 3 sin (ϕ) : 2 sin^{2} (ϕ) + 3 sin (ϕ) - 2

- 1 - (1 - 2 sin^{2} (ϕ)) + 3 sin (ϕ)

- (1 - 2 sin^{2} (ϕ)) : - 1 + 2 sin^{2} (ϕ)

- (1 - 2 sin^{2} (ϕ))

Distribute parentheses

= - (1) - (- 2 sin^{2} (ϕ))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + 2 sin^{2} (ϕ)

= - 1 - 1 + 2 sin^{2} (ϕ) + 3 sin (ϕ)

Subtract the numbers:

- 1 - 1 = - 2

= 2 sin^{2} (ϕ) + 3 sin (ϕ) - 2

= 2 sin^{2} (ϕ) + 3 sin (ϕ) - 2

- 2 + 2 sin^{2} (ϕ) + 3 sin (ϕ) = 0

Solve by substitution

- 2 + 2 sin^{2} (ϕ) + 3 sin (ϕ) = 0

Let:

sin (ϕ) = u

- 2 + 2 u^{2} + 3 u = 0

- 2 + 2 u^{2} + 3 u = 0 : u = \frac{1}{2}, u = - 2

- 2 + 2 u^{2} + 3 u = 0

Write in the standard form

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

2 u^{2} + 3 u - 2 = 0

Solve with the quadratic formula

2 u^{2} + 3 u - 2 = 0

Quadratic Equation Formula:

For

a = 2, b = 3, c = - 2

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 2 ( - 2 )}{2 \cdot 2}

3^{2} - 4 \cdot 2 (- 2) = 5

3^{2} - 4 \cdot 2 (- 2)

Apply rule

- (- a) = a

= 3^{2} + 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 3 + 5}{2 \cdot 2}, u_{2} = \frac{- 3 - 5}{2 \cdot 2}

u = \frac{- 3 + 5}{2 \cdot 2} : \frac{1}{2}

\frac{- 3 + 5}{2 \cdot 2}

Add/Subtract the numbers:

- 3 + 5 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- 3 - 5}{2 \cdot 2} : - 2

\frac{- 3 - 5}{2 \cdot 2}

Subtract the numbers:

- 3 - 5 = - 8

= \frac{- 8}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 8}{4}

Apply the fraction rule:

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

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

The solutions to the quadratic equation are:

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

Substitute back

u = sin (ϕ)

sin (ϕ) = \frac{1}{2}, sin (ϕ) = - 2

sin (ϕ) = \frac{1}{2}, sin (ϕ) = - 2

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

sin (ϕ) = \frac{1}{2}

General solutions for

sin (ϕ) = \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}

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

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

sin (ϕ) = - 2 :

No Solution

sin (ϕ) = - 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for 3sin(ϕ)-cos(2ϕ)=1 ?
The general solution for 3sin(ϕ)-cos(2ϕ)=1 is ϕ= pi/6+2pin,ϕ=(5pi)/6+2pin