What is the general solution for 2sin^2(x)-3sin(x)-1=0 ?

The general solution for 2sin^2(x)-3sin(x)-1=0 is x=-0.28460…+2pin,x=pi+0.28460…+2pin

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Popular >

2sin^2(x)-3sin(x)-1=0

Solution

2 sin^{2} (x) - 3 sin (x) - 1 = 0

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 = 196.30654 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (x) - 3 sin (x) - 1 = 0

Solve by substitution

2 sin^{2} (x) - 3 sin (x) - 1 = 0

Let:

sin (x) = u

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

2 u^{2} - 3 u - 1 = 0 : u = \frac{3 + 17}{4}, u = \frac{3 - 17}{4}

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

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

= 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}

Apply rule

- (- a) = a

= \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}

Apply rule

- (- a) = a

= \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} :

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

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

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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solvefor u,x=2tanh(u)solve for

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(x)-3sin(x)-1=0 ?
The general solution for 2sin^2(x)-3sin(x)-1=0 is x=-0.28460…+2pin,x=pi+0.28460…+2pin