What is the general solution for 2sin(θ)+cos(2θ)=1 ?

The general solution for 2sin(θ)+cos(2θ)=1 is θ=2pin,θ=pi+2pin,θ= pi/2+2pin

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2sin(θ)+cos(2θ)=1

Solution

2 sin (θ) + cos (2 θ) = 1

Solution

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

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (θ) + cos (2 θ) = 1

Subtract

1

from both sides

2 sin (θ) + cos (2 θ) - 1 = 0

Rewrite using trig identities

- 1 + cos (2 θ) + 2 sin (θ)

Use the Double Angle identity:

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

= - 1 + 1 - 2 sin^{2} (θ) + 2 sin (θ)

Simplify

= 2 sin (θ) - 2 sin^{2} (θ)

2 sin (θ) - 2 sin^{2} (θ) = 0

Solve by substitution

2 sin (θ) - 2 sin^{2} (θ) = 0

Let:

sin (θ) = u

2 u - 2 u^{2} = 0

2 u - 2 u^{2} = 0 : u = 0, u = 1

2 u - 2 u^{2} = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 2, b = 2, c = 0

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

2^{2} - 4 (- 2) \cdot 0 = 2

2^{2} - 4 (- 2) \cdot 0

Apply rule

- (- a) = a

= 2^{2} + 4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} + 0

2^{2} + 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

u_{1, 2} = \frac{- 2 \pm 2}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 2 + 2}{2 ( - 2 )}, u_{2} = \frac{- 2 - 2}{2 ( - 2 )}

u = \frac{- 2 + 2}{2 ( - 2 )} : 0

\frac{- 2 + 2}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

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

Add/Subtract the numbers:

- 2 + 2 = 0

= \frac{0}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{- 4}

Apply the fraction rule:

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

= - \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= - 0

= 0

u = \frac{- 2 - 2}{2 ( - 2 )} : 1

\frac{- 2 - 2}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

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

Subtract the numbers:

- 2 - 2 = - 4

= \frac{- 4}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 0, u = 1

Substitute back

u = sin (θ)

sin (θ) = 0, sin (θ) = 1

sin (θ) = 0, sin (θ) = 1

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

sin (θ) = 1 : θ = \frac{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{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

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