What is the general solution for sin^4(x)=sin^2(x),0<= x<2pi ?

Q: What is the general solution for sin^4(x)=sin^2(x),0<= x<2pi ?

The general solution for sin^4(x)=sin^2(x),0<= x<2pi is x= pi/2 ,x=(3pi)/2 ,x=0,x=pi

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sin^4(x)=sin^2(x),0<= x<2pi

Solution

sin^{4} (x) = sin^{2} (x), 0 \leq x < 2 π

Solution

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = π

Degrees

x = 9 0^{\circ}, x = 27 0^{\circ}, x = 0^{\circ}, x = 18 0^{\circ}

Solution steps

sin^{4} (x) = sin^{2} (x), 0 \leq x < 2 π

Solve by substitution

sin^{4} (x) = sin^{2} (x)

Let:

sin (x) = u

u^{4} = u^{2}

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

u^{4} = u^{2}

Move

u^{2}

to the left side

u^{4} = u^{2}

Subtract

u^{2}

from both sides

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

Simplify

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

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

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

v^{2} - v = 0

Solve

v^{2} - v = 0 : v = 1, v = 0

v^{2} - v = 0

Solve with the quadratic formula

v^{2} - v = 0

Quadratic Equation Formula:

For

a = 1, b = - 1, c = 0

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

v_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

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

(- 1)^{2} - 4 \cdot 1 \cdot 0

(- 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 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

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

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

v = \frac{- ( - 1 ) - 1}{2 \cdot 1} : 0

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

Apply rule

- (- a) = a

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

Subtract the numbers:

1 - 1 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

v = 1, v = 0

v = 1, v = 0

Substitute back

v = u^{2},

solve for

u

Solve

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

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

u = 0

The solutions are

u = 1, u = - 1, u = 0

Substitute back

u = sin (x)

sin (x) = 1, sin (x) = - 1, sin (x) = 0

sin (x) = 1, sin (x) = - 1, sin (x) = 0

sin (x) = 1, 0 \leq x < 2 π : x = \frac{π}{2}

sin (x) = 1, 0 \leq x < 2 π

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

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{2}

sin (x) = - 1, 0 \leq x < 2 π : x = \frac{3 π}{2}

sin (x) = - 1, 0 \leq x < 2 π

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{3 π}{2} + 2 πn

x = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{3 π}{2}

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Solutions for the range

0 \leq x < 2 π

x = 0, x = π

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}, x = 0, x = π

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Frequently Asked Questions (FAQ)

What is the general solution for sin^4(x)=sin^2(x),0<= x<2pi ?
The general solution for sin^4(x)=sin^2(x),0<= x<2pi is x= pi/2 ,x=(3pi)/2 ,x=0,x=pi