What is the general solution for 5sin^2(θ)-11sin(θ)+6=0 ?

The general solution for 5sin^2(θ)-11sin(θ)+6=0 is θ= pi/2+2pin

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Popular >

5sin^2(θ)-11sin(θ)+6=0

Solution

5 sin^{2} (θ) - 11 sin (θ) + 6 = 0

Solution

θ = \frac{π}{2} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

5 sin^{2} (θ) - 11 sin (θ) + 6 = 0

Solve by substitution

5 sin^{2} (θ) - 11 sin (θ) + 6 = 0

Let:

sin (θ) = u

5 u^{2} - 11 u + 6 = 0

5 u^{2} - 11 u + 6 = 0 : u = \frac{6}{5}, u = 1

5 u^{2} - 11 u + 6 = 0

Solve with the quadratic formula

5 u^{2} - 11 u + 6 = 0

Quadratic Equation Formula:

For

a = 5, b = - 11, c = 6

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 5 \cdot 6}{2 \cdot 5}

u_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 5 \cdot 6}{2 \cdot 5}

(- 11)^{2} - 4 \cdot 5 \cdot 6 = 1

(- 11)^{2} - 4 \cdot 5 \cdot 6

Apply exponent rule:

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

n

is even

(- 11)^{2} = 1 1^{2}

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

Multiply the numbers:

4 \cdot 5 \cdot 6 = 120

= 1 1^{2} - 120

1 1^{2} = 121

= 121 - 120

Subtract the numbers:

121 - 120 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 11 ) \pm 1}{2 \cdot 5}

Separate the solutions

u_{1} = \frac{- ( - 11 ) + 1}{2 \cdot 5}, u_{2} = \frac{- ( - 11 ) - 1}{2 \cdot 5}

u = \frac{- ( - 11 ) + 1}{2 \cdot 5} : \frac{6}{5}

\frac{- ( - 11 ) + 1}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{11 + 1}{2 \cdot 5}

Add the numbers:

11 + 1 = 12

= \frac{12}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{12}{10}

Cancel the common factor:

2

= \frac{6}{5}

u = \frac{- ( - 11 ) - 1}{2 \cdot 5} : 1

\frac{- ( - 11 ) - 1}{2 \cdot 5}

Apply rule

- (- a) = a

= \frac{11 - 1}{2 \cdot 5}

Subtract the numbers:

11 - 1 = 10

= \frac{10}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{10}{10}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = \frac{6}{5}, u = 1

Substitute back

u = sin (θ)

sin (θ) = \frac{6}{5}, sin (θ) = 1

sin (θ) = \frac{6}{5}, sin (θ) = 1

sin (θ) = \frac{6}{5} :

No Solution

sin (θ) = \frac{6}{5}

- 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

cos(x)= 11/14 ,sin(x)

cos (x) = \frac{11}{14}, sin (x)

sin^2(2x)+cos^2(3x)=1

sin^{2} (2 x) + cos^{2} (3 x) = 1

sin(θ)=0.9848

sin (θ) = 0.9848

cot^{(2)}(x)+2cot(x)+1=0

cot^{(2)} (x) + 2 cot (x) + 1 = 0

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

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

Frequently Asked Questions (FAQ)

What is the general solution for 5sin^2(θ)-11sin(θ)+6=0 ?
The general solution for 5sin^2(θ)-11sin(θ)+6=0 is θ= pi/2+2pin