What is the general solution for 6sin^2(θ)-17sin(θ)+14=-4sin(θ)+9 ?

The general solution for 6sin^2(θ)-17sin(θ)+14=-4sin(θ)+9 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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Popular >

6sin^2(θ)-17sin(θ)+14=-4sin(θ)+9

Solution

6 sin^{2} (θ) - 17 sin (θ) + 14 = - 4 sin (θ) + 9

Solution

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

Degrees

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

Solution steps

6 sin^{2} (θ) - 17 sin (θ) + 14 = - 4 sin (θ) + 9

Solve by substitution

6 sin^{2} (θ) - 17 sin (θ) + 14 = - 4 sin (θ) + 9

Let:

sin (θ) = u

6 u^{2} - 17 u + 14 = - 4 u + 9

6 u^{2} - 17 u + 14 = - 4 u + 9 : u = \frac{5}{3}, u = \frac{1}{2}

6 u^{2} - 17 u + 14 = - 4 u + 9

Move

9

to the left side

6 u^{2} - 17 u + 14 = - 4 u + 9

Subtract

9

from both sides

6 u^{2} - 17 u + 14 - 9 = - 4 u + 9 - 9

Simplify

6 u^{2} - 17 u + 5 = - 4 u

6 u^{2} - 17 u + 5 = - 4 u

Move

4 u

to the left side

6 u^{2} - 17 u + 5 = - 4 u

Add

4 u

to both sides

6 u^{2} - 17 u + 5 + 4 u = - 4 u + 4 u

Simplify

6 u^{2} - 13 u + 5 = 0

6 u^{2} - 13 u + 5 = 0

Solve with the quadratic formula

6 u^{2} - 13 u + 5 = 0

Quadratic Equation Formula:

For

a = 6, b = - 13, c = 5

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 6 \cdot 5}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 6 \cdot 5}{2 \cdot 6}

(- 13)^{2} - 4 \cdot 6 \cdot 5 = 7

(- 13)^{2} - 4 \cdot 6 \cdot 5

Apply exponent rule:

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

n

is even

(- 13)^{2} = 1 3^{2}

= 1 3^{2} - 4 \cdot 6 \cdot 5

Multiply the numbers:

4 \cdot 6 \cdot 5 = 120

= 1 3^{2} - 120

1 3^{2} = 169

= 169 - 120

Subtract the numbers:

169 - 120 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 13 ) \pm 7}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 13 ) + 7}{2 \cdot 6}, u_{2} = \frac{- ( - 13 ) - 7}{2 \cdot 6}

u = \frac{- ( - 13 ) + 7}{2 \cdot 6} : \frac{5}{3}

\frac{- ( - 13 ) + 7}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{13 + 7}{2 \cdot 6}

Add the numbers:

13 + 7 = 20

= \frac{20}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{20}{12}

Cancel the common factor:

4

= \frac{5}{3}

u = \frac{- ( - 13 ) - 7}{2 \cdot 6} : \frac{1}{2}

\frac{- ( - 13 ) - 7}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{13 - 7}{2 \cdot 6}

Subtract the numbers:

13 - 7 = 6

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{6}{12}

Cancel the common factor:

6

= \frac{1}{2}

The solutions to the quadratic equation are:

u = \frac{5}{3}, u = \frac{1}{2}

Substitute back

u = sin (θ)

sin (θ) = \frac{5}{3}, sin (θ) = \frac{1}{2}

sin (θ) = \frac{5}{3}, sin (θ) = \frac{1}{2}

sin (θ) = \frac{5}{3} :

No Solution

sin (θ) = \frac{5}{3}

- 1 \leq sin (x) \leq 1

No Solution

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}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

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cos (x + π) - cos (x) - 1 = 0

sin(x)= 15/17

sin (x) = \frac{15}{17}

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2 cos (x) - sec (x) = 1

cos(x)tan(x)=5tan(x)

cos (x) tan (x) = 5 tan (x)

Frequently Asked Questions (FAQ)

What is the general solution for 6sin^2(θ)-17sin(θ)+14=-4sin(θ)+9 ?
The general solution for 6sin^2(θ)-17sin(θ)+14=-4sin(θ)+9 is θ= pi/6+2pin,θ=(5pi)/6+2pin