What is the general solution for 7sin^2(θ)-15sin(θ)+8=0 ?

The general solution for 7sin^2(θ)-15sin(θ)+8=0 is θ= pi/2+2pin

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Popular >

7sin^2(θ)-15sin(θ)+8=0

Solution

7 sin^{2} (θ) - 15 sin (θ) + 8 = 0

Solution

θ = \frac{π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

7 sin^{2} (θ) - 15 sin (θ) + 8 = 0

Solve by substitution

7 sin^{2} (θ) - 15 sin (θ) + 8 = 0

Let:

sin (θ) = u

7 u^{2} - 15 u + 8 = 0

7 u^{2} - 15 u + 8 = 0 : u = \frac{8}{7}, u = 1

7 u^{2} - 15 u + 8 = 0

Solve with the quadratic formula

7 u^{2} - 15 u + 8 = 0

Quadratic Equation Formula:

For

a = 7, b = - 15, c = 8

u_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 7 \cdot 8}{2 \cdot 7}

u_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 7 \cdot 8}{2 \cdot 7}

(- 15)^{2} - 4 \cdot 7 \cdot 8 = 1

(- 15)^{2} - 4 \cdot 7 \cdot 8

Apply exponent rule:

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

n

is even

(- 15)^{2} = 1 5^{2}

= 1 5^{2} - 4 \cdot 7 \cdot 8

Multiply the numbers:

4 \cdot 7 \cdot 8 = 224

= 1 5^{2} - 224

1 5^{2} = 225

= 225 - 224

Subtract the numbers:

225 - 224 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 15 ) \pm 1}{2 \cdot 7}

Separate the solutions

u_{1} = \frac{- ( - 15 ) + 1}{2 \cdot 7}, u_{2} = \frac{- ( - 15 ) - 1}{2 \cdot 7}

u = \frac{- ( - 15 ) + 1}{2 \cdot 7} : \frac{8}{7}

\frac{- ( - 15 ) + 1}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{15 + 1}{2 \cdot 7}

Add the numbers:

15 + 1 = 16

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

Multiply the numbers:

2 \cdot 7 = 14

= \frac{16}{14}

Cancel the common factor:

2

= \frac{8}{7}

u = \frac{- ( - 15 ) - 1}{2 \cdot 7} : 1

\frac{- ( - 15 ) - 1}{2 \cdot 7}

Apply rule

- (- a) = a

= \frac{15 - 1}{2 \cdot 7}

Subtract the numbers:

15 - 1 = 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{8}{7}, u = 1

Substitute back

u = sin (θ)

sin (θ) = \frac{8}{7}, sin (θ) = 1

sin (θ) = \frac{8}{7}, sin (θ) = 1

sin (θ) = \frac{8}{7} :

No Solution

sin (θ) = \frac{8}{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

Popular Examples

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sin (5 t) = 0

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cos (x) = 0, - π \leq x \leq π

sin^2(x)+cos(x)=2

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

sin(θ)=-4/5 ,(7pi)/2 <= θ<4pi,cos(2θ)

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

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

Frequently Asked Questions (FAQ)

What is the general solution for 7sin^2(θ)-15sin(θ)+8=0 ?
The general solution for 7sin^2(θ)-15sin(θ)+8=0 is θ= pi/2+2pin