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Popular Trigonometry >

-3cos^2(θ)-sin(θ)+3=sin(θ)

Solution

- 3 cos^{2} (θ) - sin (θ) + 3 = sin (θ)

Solution

θ = 0.72972 \dots + 2 πn, θ = π - 0.72972 \dots + 2 πn, θ = 2 πn, θ = π + 2 πn

Degrees

θ = 41.81031 \dots^{\circ} + 36 0^{\circ} n, θ = 138.18968 \dots^{\circ} + 36 0^{\circ} n, θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

- 3 cos^{2} (θ) - sin (θ) + 3 = sin (θ)

Subtract

sin (θ)

from both sides

- 3 cos^{2} (θ) - 2 sin (θ) + 3 = 0

Rewrite using trig identities

3 - 2 sin (θ) - 3 cos^{2} (θ)

Use the Pythagorean identity:

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

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

= 3 - 2 sin (θ) - 3 (1 - sin^{2} (θ))

Simplify

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

3 - 2 sin (θ) - 3 (1 - sin^{2} (θ))

Expand

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

- 3 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = 1, c = sin^{2} (θ)

= - 3 \cdot 1 - (- 3) sin^{2} (θ)

Apply minus-plus rules

- (- a) = a

= - 3 \cdot 1 + 3 sin^{2} (θ)

Multiply the numbers:

3 \cdot 1 = 3

= - 3 + 3 sin^{2} (θ)

= 3 - 2 sin (θ) - 3 + 3 sin^{2} (θ)

Simplify

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

3 - 2 sin (θ) - 3 + 3 sin^{2} (θ)

Group like terms

= - 2 sin (θ) + 3 sin^{2} (θ) + 3 - 3

3 - 3 = 0

= 3 sin^{2} (θ) - 2 sin (θ)

= 3 sin^{2} (θ) - 2 sin (θ)

= 3 sin^{2} (θ) - 2 sin (θ)

- 2 sin (θ) + 3 sin^{2} (θ) = 0

Solve by substitution

- 2 sin (θ) + 3 sin^{2} (θ) = 0

Let:

sin (θ) = u

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

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

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

Write in the standard form

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

3 u^{2} - 2 u = 0

Solve with the quadratic formula

3 u^{2} - 2 u = 0

Quadratic Equation Formula:

For

a = 3, b = - 2, c = 0

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

u_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

(- 2)^{2} - 4 \cdot 3 \cdot 0 = 2

(- 2)^{2} - 4 \cdot 3 \cdot 0

Apply exponent rule:

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

n

is even

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

= 2^{2} - 4 \cdot 3 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a,

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

2 + 2 = 4

= \frac{4}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{4}{6}

Cancel the common factor:

2

= \frac{2}{3}

u = \frac{- ( - 2 ) - 2}{2 \cdot 3} : 0

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

Apply rule

- (- a) = a

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

Subtract the numbers:

2 - 2 = 0

= \frac{0}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{0}{6}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = \frac{2}{3}, u = 0

Substitute back

u = sin (θ)

sin (θ) = \frac{2}{3}, sin (θ) = 0

sin (θ) = \frac{2}{3}, sin (θ) = 0

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

sin (θ) = \frac{2}{3}

Apply trig inverse properties

sin (θ) = \frac{2}{3}

General solutions for

sin (θ) = \frac{2}{3}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{2}{3}) + 2 πn, θ = π - arcsin (\frac{2}{3}) + 2 πn

θ = arcsin (\frac{2}{3}) + 2 πn, θ = π - arcsin (\frac{2}{3}) + 2 πn

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

Combine all the solutions

θ = arcsin (\frac{2}{3}) + 2 πn, θ = π - arcsin (\frac{2}{3}) + 2 πn, θ = 2 πn, θ = π + 2 πn

Show solutions in decimal form

θ = 0.72972 \dots + 2 πn, θ = π - 0.72972 \dots + 2 πn, θ = 2 πn, θ = π + 2 πn

Graph

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