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

The general solution for sin^2(x)-3sin(x)-2=0 is x=-0.59626…+2pin,x=pi+0.59626…+2pin

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Popular >

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

Solution

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

Solution

x = - 0.59626 \dots + 2 πn, x = π + 0.59626 \dots + 2 πn

Degrees

x = - 34.16325 \dots^{\circ} + 36 0^{\circ} n, x = 214.16325 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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) = 17

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 2 = 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 1}

Separate the solutions

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

u = \frac{- ( - 3 ) + 17}{2 \cdot 1} : \frac{3 + 17}{2}

\frac{- ( - 3 ) + 17}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 + 17}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{3 + 17}{2}

u = \frac{- ( - 3 ) - 17}{2 \cdot 1} : \frac{3 - 17}{2}

\frac{- ( - 3 ) - 17}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 - 17}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{3 - 17}{2}

The solutions to the quadratic equation are:

u = \frac{3 + 17}{2}, u = \frac{3 - 17}{2}

Substitute back

u = sin (x)

sin (x) = \frac{3 + 17}{2}, sin (x) = \frac{3 - 17}{2}

sin (x) = \frac{3 + 17}{2}, sin (x) = \frac{3 - 17}{2}

sin (x) = \frac{3 + 17}{2} :

No Solution

sin (x) = \frac{3 + 17}{2}

- 1 \leq sin (x) \leq 1

No Solution

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

sin (x) = \frac{3 - 17}{2}

Apply trig inverse properties

sin (x) = \frac{3 - 17}{2}

General solutions for

sin (x) = \frac{3 - 17}{2}

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

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

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

Combine all the solutions

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

Show solutions in decimal form

x = - 0.59626 \dots + 2 πn, x = π + 0.59626 \dots + 2 πn

Popular Examples

-11tan(x)+1=-2(4+tan(x))

- 11 tan (x) + 1 = - 2 (4 + tan (x))

16=-10cos((2pi)/(150)t)+10

16 = - 10 cos (\frac{2 π}{150} t) + 10

2cos(x)-sqrt(3)=0,0<= x<= 2pi

2 cos (x) - 3 = 0, 0 \leq x \leq 2 π

sec^2(x)-1=0,0<= x<= 2pi

sec^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

3cos(x/2)+sqrt(3)=cos(x/2)

3 cos (\frac{x}{2}) + 3 = cos (\frac{x}{2})

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)-3sin(x)-2=0 ?
The general solution for sin^2(x)-3sin(x)-2=0 is x=-0.59626…+2pin,x=pi+0.59626…+2pin