What is the general solution for sin(x)-sin^2(x)= 1/4 ?

The general solution for sin(x)-sin^2(x)= 1/4 is x= pi/6+2pin,x=(5pi)/6+2pin

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sin(x)-sin^2(x)= 1/4

Solution

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

Solution

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

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

sin (x) = u

u - u^{2} = \frac{1}{4}

u - u^{2} = \frac{1}{4} : u = \frac{1}{2}

u - u^{2} = \frac{1}{4}

Multiply both sides by

4

u - u^{2} = \frac{1}{4}

Multiply both sides by

4

u \cdot 4 - u^{2} \cdot 4 = \frac{1}{4} \cdot 4

Simplify

4 u - 4 u^{2} = 1

4 u - 4 u^{2} = 1

Move

1

to the left side

4 u - 4 u^{2} = 1

Subtract

1

from both sides

4 u - 4 u^{2} - 1 = 1 - 1

Simplify

4 u - 4 u^{2} - 1 = 0

4 u - 4 u^{2} - 1 = 0

Write in the standard form

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

- 4 u^{2} + 4 u - 1 = 0

Solve with the quadratic formula

- 4 u^{2} + 4 u - 1 = 0

Quadratic Equation Formula:

For

a = - 4, b = 4, c = - 1

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

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

4^{2} - 4 (- 4) (- 1) = 0

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

Apply rule

- (- a) = a

= 4^{2} - 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

u_{1, 2} = \frac{- 4 \pm 0}{2 ( - 4 )}

u = \frac{- 4}{2 ( - 4 )}

\frac{- 4}{2 ( - 4 )} = \frac{1}{2}

\frac{- 4}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{1}{2}

The solution to the quadratic equation is:

u = \frac{1}{2}

Substitute back

u = sin (x)

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

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

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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Popular Examples

20sin(x)=20cos(x)

20 sin (x) = 20 cos (x)

(-5cot(x))/2+7/3 =-1/6

\frac{- 5 cot ( x )}{2} + \frac{7}{3} = - \frac{1}{6}

8.59tan^2(θ)-24tan(θ)+15.29=0

8.59 tan^{2} (θ) - 24 tan (θ) + 15.29 = 0

(csc(x))=-csc(x)cot(x)

(csc (x)) = - csc (x) cot (x)

cot(x)=-4/3

cot (x) = - \frac{4}{3}

Frequently Asked Questions (FAQ)

What is the general solution for sin(x)-sin^2(x)= 1/4 ?
The general solution for sin(x)-sin^2(x)= 1/4 is x= pi/6+2pin,x=(5pi)/6+2pin