What is the general solution for sin^4(x)+sin^2(x)=sin^6(x) ?

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

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Popular Trigonometry >

sin^4(x)+sin^2(x)=sin^6(x)

Solution

sin^{4} (x) + sin^{2} (x) = sin^{6} (x)

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{4} (x) + sin^{2} (x) = sin^{6} (x)

Solve by substitution

sin^{4} (x) + sin^{2} (x) = sin^{6} (x)

Let:

sin (x) = u

u^{4} + u^{2} = u^{6}

u^{4} + u^{2} = u^{6} : u = 0, u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}, u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

u^{4} + u^{2} = u^{6}

Switch sides

u^{6} = u^{4} + u^{2}

Move

u^{2}

to the left side

u^{6} = u^{4} + u^{2}

Subtract

u^{2}

from both sides

u^{6} - u^{2} = u^{4} + u^{2} - u^{2}

Simplify

u^{6} - u^{2} = u^{4}

u^{6} - u^{2} = u^{4}

Move

u^{4}

to the left side

u^{6} - u^{2} = u^{4}

Subtract

u^{4}

from both sides

u^{6} - u^{2} - u^{4} = u^{4} - u^{4}

Simplify

u^{6} - u^{2} - u^{4} = 0

u^{6} - u^{2} - u^{4} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{6} - u^{4} - u^{2} = 0

Rewrite the equation with

v = u^{2}, v^{2} = u^{4}

and

v^{3} = u^{6}

v^{3} - v^{2} - v = 0

Solve

v^{3} - v^{2} - v = 0 : v = 0, v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

v^{3} - v^{2} - v = 0

Factor

v^{3} - v^{2} - v : v (v^{2} - v - 1)

v^{3} - v^{2} - v

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

v^{2} = vv

= v^{2} v - vv - v

Factor out common term

v

= v (v^{2} - v - 1)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

v = 0 or v^{2} - v - 1 = 0

Solve

v^{2} - v - 1 = 0 : v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

v^{2} - v - 1 = 0

Solve with the quadratic formula

v^{2} - v - 1 = 0

Quadratic Equation Formula:

For

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

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

v_{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

- (- a) = a

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

(- 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 1 = 4

4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 4

= 1 + 4

Add the numbers:

1 + 4 = 5

= 5

v_{1, 2} = \frac{- ( - 1 ) \pm 5}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 1 ) + 5}{2 \cdot 1}, v_{2} = \frac{- ( - 1 ) - 5}{2 \cdot 1}

v = \frac{- ( - 1 ) + 5}{2 \cdot 1} : \frac{1 + 5}{2}

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 + 5}{2}

v = \frac{- ( - 1 ) - 5}{2 \cdot 1} : \frac{1 - 5}{2}

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

Apply rule

- (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{1 - 5}{2}

The solutions to the quadratic equation are:

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

The solutions are

v = 0, v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

v = 0, v = \frac{1 + 5}{2}, v = \frac{1 - 5}{2}

Substitute back

v = u^{2},

solve for

u

Solve

u^{2} = 0 : u = 0

u^{2} = 0

Apply rule

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

u = 0

Solve

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

u^{2} = \frac{1 + 5}{2}

For

x^{2} = f (a)

the solutions are

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

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

Solve

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

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

For

x^{2} = f (a)

the solutions are

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

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

The solutions are

u = 0, u = \frac{1 + 5}{2}, u = - \frac{1 + 5}{2}, u = \frac{1 - 5}{2}, u = - \frac{1 - 5}{2}

Substitute back

u = sin (x)

sin (x) = 0, 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) = 0, 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) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

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

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

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} : 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

Combine all the solutions

x = 2 πn, x = π + 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

Since the equation is undefined for:

arcsin \frac{1 - 5}{2} + 2 πn, π + arcsin - \frac{1 - 5}{2} + 2 πn, arcsin - \frac{1 - 5}{2} + 2 πn, π + arcsin \frac{1 - 5}{2} + 2 πn

x = 2 πn, x = π + 2 πn

Graph

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Frequently Asked Questions (FAQ)

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