What is the general solution for 2sin^2(x)+3sin(x)=1 ?

The general solution for 2sin^2(x)+3sin(x)=1 is x=0.28460…+2pin,x=pi-0.28460…+2pin

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Popular >

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

Solution

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

Solution

x = 0.28460 \dots + 2 πn, x = π - 0.28460 \dots + 2 πn

Degrees

x = 16.30654 \dots^{\circ} + 36 0^{\circ} n, x = 163.69345 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

2 u^{2} + 3 u = 1

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

2 u^{2} + 3 u = 1

Move

1

to the left side

2 u^{2} + 3 u = 1

Subtract

1

from both sides

2 u^{2} + 3 u - 1 = 1 - 1

Simplify

2 u^{2} + 3 u - 1 = 0

2 u^{2} + 3 u - 1 = 0

Solve with the quadratic formula

2 u^{2} + 3 u - 1 = 0

Quadratic Equation Formula:

For

a = 2, b = 3, c = - 1

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

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

3^{2} - 4 \cdot 2 (- 1) = 17

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 3^{2} + 8

3^{2} = 9

= 9 + 8

Add the numbers:

9 + 8 = 17

= 17

u_{1, 2} = \frac{- 3 \pm 17}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 3 + 17}{2 \cdot 2}, u_{2} = \frac{- 3 - 17}{2 \cdot 2}

u = \frac{- 3 + 17}{2 \cdot 2} : \frac{- 3 + 17}{4}

\frac{- 3 + 17}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 3 + 17}{4}

u = \frac{- 3 - 17}{2 \cdot 2} : \frac{- 3 - 17}{4}

\frac{- 3 - 17}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 3 - 17}{4}

The solutions to the quadratic equation are:

u = \frac{- 3 + 17}{4}, u = \frac{- 3 - 17}{4}

Substitute back

u = sin (x)

sin (x) = \frac{- 3 + 17}{4}, sin (x) = \frac{- 3 - 17}{4}

sin (x) = \frac{- 3 + 17}{4}, sin (x) = \frac{- 3 - 17}{4}

sin (x) = \frac{- 3 + 17}{4} : x = arcsin (\frac{- 3 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 3 + 17}{4}) + 2 πn

sin (x) = \frac{- 3 + 17}{4}

Apply trig inverse properties

sin (x) = \frac{- 3 + 17}{4}

General solutions for

sin (x) = \frac{- 3 + 17}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{- 3 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 3 + 17}{4}) + 2 πn

x = arcsin (\frac{- 3 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 3 + 17}{4}) + 2 πn

sin (x) = \frac{- 3 - 17}{4} :

No Solution

sin (x) = \frac{- 3 - 17}{4}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (\frac{- 3 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 3 + 17}{4}) + 2 πn

Show solutions in decimal form

x = 0.28460 \dots + 2 πn, x = π - 0.28460 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)+3sin(x)=1 ?
The general solution for 2sin^2(x)+3sin(x)=1 is x=0.28460…+2pin,x=pi-0.28460…+2pin