What is the general solution for 7sin^2(θ)-16sin(θ)+9=0 ?

The general solution for 7sin^2(θ)-16sin(θ)+9=0 is θ= pi/2+2pin

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Popular >

7sin^2(θ)-16sin(θ)+9=0

Solution

7 sin^{2} (θ) - 16 sin (θ) + 9 = 0

Solution

θ = \frac{π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

7 sin^{2} (θ) - 16 sin (θ) + 9 = 0

Solve by substitution

7 sin^{2} (θ) - 16 sin (θ) + 9 = 0

Let:

sin (θ) = u

7 u^{2} - 16 u + 9 = 0

7 u^{2} - 16 u + 9 = 0 : u = \frac{9}{7}, u = 1

7 u^{2} - 16 u + 9 = 0

Solve with the quadratic formula

7 u^{2} - 16 u + 9 = 0

Quadratic Equation Formula:

For

a = 7, b = - 16, c = 9

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 7 \cdot 9}{2 \cdot 7}

u_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 7 \cdot 9}{2 \cdot 7}

(- 16)^{2} - 4 \cdot 7 \cdot 9 = 2

(- 16)^{2} - 4 \cdot 7 \cdot 9

Apply exponent rule:

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

n

is even

(- 16)^{2} = 1 6^{2}

= 1 6^{2} - 4 \cdot 7 \cdot 9

Multiply the numbers:

4 \cdot 7 \cdot 9 = 252

= 1 6^{2} - 252

1 6^{2} = 256

= 256 - 252

Subtract the numbers:

256 - 252 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 16 ) \pm 2}{2 \cdot 7}

Separate the solutions

u_{1} = \frac{- ( - 16 ) + 2}{2 \cdot 7}, u_{2} = \frac{- ( - 16 ) - 2}{2 \cdot 7}

u = \frac{- ( - 16 ) + 2}{2 \cdot 7} : \frac{9}{7}

\frac{- ( - 16 ) + 2}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{16 + 2}{2 \cdot 7}

Add the numbers:

16 + 2 = 18

= \frac{18}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{18}{14}

Cancel the common factor:

2

= \frac{9}{7}

u = \frac{- ( - 16 ) - 2}{2 \cdot 7} : 1

\frac{- ( - 16 ) - 2}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{16 - 2}{2 \cdot 7}

Subtract the numbers:

16 - 2 = 14

= \frac{14}{2 \cdot 7}

Multiply the numbers:

2 \cdot 7 = 14

= \frac{14}{14}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = \frac{9}{7}, u = 1

Substitute back

u = sin (θ)

sin (θ) = \frac{9}{7}, sin (θ) = 1

sin (θ) = \frac{9}{7}, sin (θ) = 1

sin (θ) = \frac{9}{7} :

No Solution

sin (θ) = \frac{9}{7}

- 1 \leq sin (x) \leq 1

No Solution

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 = \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 7sin^2(θ)-16sin(θ)+9=0 ?
The general solution for 7sin^2(θ)-16sin(θ)+9=0 is θ= pi/2+2pin