What is the general solution for sin^2(x)+3sin(x)+7=5 ?

The general solution for sin^2(x)+3sin(x)+7=5 is x=(3pi)/2+2pin

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Popular >

sin^2(x)+3sin(x)+7=5

Solution

sin^{2} (x) + 3 sin (x) + 7 = 5

Solution

x = \frac{3 π}{2} + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (x) + 3 sin (x) + 7 = 5

Solve by substitution

sin^{2} (x) + 3 sin (x) + 7 = 5

Let:

sin (x) = u

u^{2} + 3 u + 7 = 5

u^{2} + 3 u + 7 = 5 : u = - 1, u = - 2

u^{2} + 3 u + 7 = 5

Move

5

to the left side

u^{2} + 3 u + 7 = 5

Subtract

5

from both sides

u^{2} + 3 u + 7 - 5 = 5 - 5

Simplify

u^{2} + 3 u + 2 = 0

u^{2} + 3 u + 2 = 0

Solve with the quadratic formula

u^{2} + 3 u + 2 = 0

Quadratic Equation Formula:

For

a = 1, b = 3, c = 2

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

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

3^{2} - 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 3^{2} - 8

3^{2} = 9

= 9 - 8

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 3 \pm 1}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 3 + 1}{2 \cdot 1}, u_{2} = \frac{- 3 - 1}{2 \cdot 1}

u = \frac{- 3 + 1}{2 \cdot 1} : - 1

\frac{- 3 + 1}{2 \cdot 1}

Add/Subtract the numbers:

- 3 + 1 = - 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 2}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

u = \frac{- 3 - 1}{2 \cdot 1} : - 2

\frac{- 3 - 1}{2 \cdot 1}

Subtract the numbers:

- 3 - 1 = - 4

= \frac{- 4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = - 1, u = - 2

Substitute back

u = sin (x)

sin (x) = - 1, sin (x) = - 2

sin (x) = - 1, sin (x) = - 2

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

General solutions for

sin (x) = - 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}

x = \frac{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

sin (x) = - 2 :

No Solution

sin (x) = - 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = \frac{3 π}{2} + 2 πn

Popular Examples

cos(2x)=-0.6,-180<= x<= 180

cos (2 x) = - 0.6, - 18 0^{\circ} \leq x \leq 18 0^{\circ}

2sqrt(3)cos(θ/2)=-3

23 cos (\frac{θ}{2}) = - 3

1/(cos^2(θ/2))=12cos(θ)

\frac{1}{cos ^{2} ( \frac{θ}{2} )} = 12 cos (θ)

sqrt(3)cos(x)-sin(x)=sqrt(2)

3 cos (x) - sin (x) = 2

cos(x)=(sqrt(34))/(sqrt(70))

cos (x) = \frac{34}{70}

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)+3sin(x)+7=5 ?
The general solution for sin^2(x)+3sin(x)+7=5 is x=(3pi)/2+2pin