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Popular >

6sin^2(θ)-5sin(θ)=-1

Solution

6 sin^{2} (θ) - 5 sin (θ) = - 1

Solution

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = 0.33983 \dots + 2 πn, θ = π - 0.33983 \dots + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n, θ = 19.47122 \dots^{\circ} + 36 0^{\circ} n, θ = 160.52877 \dots^{\circ} + 36 0^{\circ} n

Solution steps

6 sin^{2} (θ) - 5 sin (θ) = - 1

Solve by substitution

6 sin^{2} (θ) - 5 sin (θ) = - 1

Let:

sin (θ) = u

6 u^{2} - 5 u = - 1

6 u^{2} - 5 u = - 1 : u = \frac{1}{2}, u = \frac{1}{3}

6 u^{2} - 5 u = - 1

Move

1

to the left side

6 u^{2} - 5 u = - 1

Add

1

to both sides

6 u^{2} - 5 u + 1 = - 1 + 1

Simplify

6 u^{2} - 5 u + 1 = 0

6 u^{2} - 5 u + 1 = 0

Solve with the quadratic formula

6 u^{2} - 5 u + 1 = 0

Quadratic Equation Formula:

For

a = 6, b = - 5, c = 1

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

(- 5)^{2} - 4 \cdot 6 \cdot 1 = 1

(- 5)^{2} - 4 \cdot 6 \cdot 1

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 6}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 6}

u = \frac{- ( - 5 ) + 1}{2 \cdot 6} : \frac{1}{2}

\frac{- ( - 5 ) + 1}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{5 + 1}{2 \cdot 6}

Add the numbers:

5 + 1 = 6

= \frac{6}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{6}{12}

Cancel the common factor:

6

= \frac{1}{2}

u = \frac{- ( - 5 ) - 1}{2 \cdot 6} : \frac{1}{3}

\frac{- ( - 5 ) - 1}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{5 - 1}{2 \cdot 6}

Subtract the numbers:

5 - 1 = 4

= \frac{4}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{4}{12}

Cancel the common factor:

4

= \frac{1}{3}

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = \frac{1}{3}

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{2}, sin (θ) = \frac{1}{3}

sin (θ) = \frac{1}{2}, sin (θ) = \frac{1}{3}

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

sin (θ) = \frac{1}{3}

Apply trig inverse properties

sin (θ) = \frac{1}{3}

General solutions for

sin (θ) = \frac{1}{3}

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

θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn

θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = arcsin (\frac{1}{3}) + 2 πn, θ = π - arcsin (\frac{1}{3}) + 2 πn

Show solutions in decimal form

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = 0.33983 \dots + 2 πn, θ = π - 0.33983 \dots + 2 πn

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