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

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

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3(1-sin(θ))=2cos^2(θ)

Solution

3 (1 - sin (θ)) = 2 cos^{2} (θ)

Solution

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

3 (1 - sin (θ)) = 2 cos^{2} (θ)

Subtract

2 cos^{2} (θ)

from both sides

3 (1 - sin (θ)) - 2 cos^{2} (θ) = 0

Rewrite using trig identities

(1 - sin (θ)) \cdot 3 - 2 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= (1 - sin (θ)) \cdot 3 - 2 (1 - sin^{2} (θ))

Simplify

(1 - sin (θ)) \cdot 3 - 2 (1 - sin^{2} (θ)) : 2 sin^{2} (θ) - 3 sin (θ) + 1

(1 - sin (θ)) \cdot 3 - 2 (1 - sin^{2} (θ))

= 3 (1 - sin (θ)) - 2 (1 - sin^{2} (θ))

Expand

3 (1 - sin (θ)) : 3 - 3 sin (θ)

3 (1 - sin (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = sin (θ)

= 3 \cdot 1 - 3 sin (θ)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 sin (θ)

= 3 - 3 sin (θ) - 2 (1 - sin^{2} (θ))

Expand

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

- 2 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 2, b = 1, c = sin^{2} (θ)

= - 2 \cdot 1 - (- 2) sin^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 2 \cdot 1 + 2 sin^{2} (θ)

Multiply the numbers:

2 \cdot 1 = 2

= - 2 + 2 sin^{2} (θ)

= 3 - 3 sin (θ) - 2 + 2 sin^{2} (θ)

Simplify

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

3 - 3 sin (θ) - 2 + 2 sin^{2} (θ)

Group like terms

= - 3 sin (θ) + 2 sin^{2} (θ) + 3 - 2

Add/Subtract the numbers:

3 - 2 = 1

= 2 sin^{2} (θ) - 3 sin (θ) + 1

= 2 sin^{2} (θ) - 3 sin (θ) + 1

= 2 sin^{2} (θ) - 3 sin (θ) + 1

1 + 2 sin^{2} (θ) - 3 sin (θ) = 0

Solve by substitution

1 + 2 sin^{2} (θ) - 3 sin (θ) = 0

Let:

sin (θ) = u

1 + 2 u^{2} - 3 u = 0

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

1 + 2 u^{2} - 3 u = 0

Write in the standard form

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

2 u^{2} - 3 u + 1 = 0

Solve with the quadratic formula

2 u^{2} - 3 u + 1 = 0

Quadratic Equation Formula:

For

a = 2, b = - 3, c = 1

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

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

(- 3)^{2} - 4 \cdot 2 \cdot 1 = 1

(- 3)^{2} - 4 \cdot 2 \cdot 1

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} - 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

\frac{- ( - 3 ) + 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 + 1}{2 \cdot 2}

Add the numbers:

3 + 1 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

u = \frac{- ( - 3 ) - 1}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 3 ) - 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 - 1}{2 \cdot 2}

Subtract the numbers:

3 - 1 = 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (θ)

sin (θ) = 1, sin (θ) = \frac{1}{2}

sin (θ) = 1, sin (θ) = \frac{1}{2}

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

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

Combine all the solutions

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

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tan (θ) = 0.11246

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Frequently Asked Questions (FAQ)

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