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

The general solution for 2sin^2(x)-3sin(x)=-1 is x= pi/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin

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

Solution

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

Solution

x = \frac{π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\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 = 1, u = \frac{1}{2}

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

Move

1

to the left side

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

Add

1

to 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 \cdot 1}{2 \cdot 2}

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

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

(- 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

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

3 + 1 = 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 1 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

u = sin (x)

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

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

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \frac{1}{2}

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Popular Examples

3cos^2(θ)+2sin(θ)-1=0

3 cos^{2} (θ) + 2 sin (θ) - 1 = 0

2sin^2(θ)-3sin(θ)-2=0

2 sin^{2} (θ) - 3 sin (θ) - 2 = 0

sqrt(2)cos(x)-1=0

2 cos (x) - 1 = 0

cos(x)=cos(2x)

cos (x) = cos (2 x)

0=sin(x)

0 = sin (x)

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= pi/2+2pin,x= pi/6+2pin,x=(5pi)/6+2pin