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

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

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

Solution

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

Solution

x = \frac{π}{2} + 2 πn

Degrees

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

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

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

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

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

Move

1

to the left side

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

Add

1

to both sides

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

Simplify

u^{2} + 2 = 3 u

u^{2} + 2 = 3 u

Move

3 u

to the left side

u^{2} + 2 = 3 u

Subtract

3 u

from both sides

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

Simplify

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

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

Write in the standard form

a x^{2} + b x + c = 0

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 = 1

(- 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

Subtract the numbers:

9 - 8 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

3 + 1 = 4

= \frac{4}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

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

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

Apply rule

- (- a) = a

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

Subtract the numbers:

3 - 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = 2, u = 1

Substitute back

u = sin (x)

sin (x) = 2, sin (x) = 1

sin (x) = 2, sin (x) = 1

sin (x) = 2 :

No Solution

sin (x) = 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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tan(θ)= 14/15

tan (θ) = \frac{14}{15}

sin(2x)+cos(2x)=-1

sin (2 x) + cos (2 x) = - 1

sin(-θ)=-3/5 ,tan(θ)>0

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

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

Frequently Asked Questions (FAQ)

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