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

The general solution for sin(x)+sin^2(x)=1 is x=0.66623…+2pin,x=pi-0.66623…+2pin

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Popular >

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

Solution

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

Solution

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

Degrees

x = 38.17270 \dots^{\circ} + 36 0^{\circ} n, x = 141.82729 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

u + u^{2} = 1

u + u^{2} = 1 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

u + u^{2} = 1

Move

1

to the left side

u + u^{2} = 1

Subtract

1

from both sides

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

Simplify

u + u^{2} - 1 = 0

u + u^{2} - 1 = 0

Write in the standard form

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

u^{2} + u - 1 = 0

Solve with the quadratic formula

u^{2} + u - 1 = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = - 1

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

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

Separate the solutions

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

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

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

Multiply the numbers:

2 \cdot 1 = 2

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

The solutions to the quadratic equation are:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

Substitute back

u = sin (x)

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

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

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

sin (x) = \frac{- 1 + 5}{2}

Apply trig inverse properties

sin (x) = \frac{- 1 + 5}{2}

General solutions for

sin (x) = \frac{- 1 + 5}{2}

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

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

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

No Solution

sin (x) = \frac{- 1 - 5}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

x = arcsin (\frac{- 1 + 5}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 5}{2}) + 2 πn

Show solutions in decimal form

x = 0.66623 \dots + 2 πn, x = π - 0.66623 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)+sin^2(x)=1 ?
The general solution for sin(x)+sin^2(x)=1 is x=0.66623…+2pin,x=pi-0.66623…+2pin