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

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

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Popular >

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

Solution

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

Solution

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

(- 1)^{2} - 4 (- 2) \cdot 3 = 5

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

Apply rule

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 3 = 24

4 \cdot 2 \cdot 3

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 24

= 1 + 24

Add the numbers:

1 + 24 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

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

Separate the solutions

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

u = \frac{- ( - 1 ) + 5}{2 ( - 2 )} : - \frac{3}{2}

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Add the numbers:

1 + 5 = 6

= \frac{6}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{6}{- 4}

Apply the fraction rule:

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

= - \frac{6}{4}

Cancel the common factor:

2

= - \frac{3}{2}

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

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

Remove parentheses:

(- a) = - a, - (- a) = a

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

Subtract the numbers:

1 - 5 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - \frac{3}{2}, u = 1

Substitute back

u = sin (x)

sin (x) = - \frac{3}{2}, sin (x) = 1

sin (x) = - \frac{3}{2}, sin (x) = 1

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = 1 : x = \frac{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{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn

Popular Examples

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3 tan (x) - cot (x) = 0

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sin^{2} (x) = \frac{( 2 m - 1 )}{4}

cos(2x)-0.25=0

cos (2 x) - 0.25 = 0

2tan^2(x)+1=cos(x)

2 tan^{2} (x) + 1 = cos (x)

Frequently Asked Questions (FAQ)

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