What is the general solution for sin^2(6x)+sin^2(3x)=0 ?

The general solution for sin^2(6x)+sin^2(3x)=0 is x=(2pin)/3 ,x=(pi+2pin)/3

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Popular Trigonometry >

sin^2(6x)+sin^2(3x)=0

Solution

sin^{2} (6 x) + sin^{2} (3 x) = 0

Solution

x = \frac{2 πn}{3}, x = \frac{π + 2 πn}{3}

Degrees

x = 0^{\circ} + 12 0^{\circ} n, x = 6 0^{\circ} + 12 0^{\circ} n

Solution steps

sin^{2} (6 x) + sin^{2} (3 x) = 0

Let:

u = 3 x

sin^{2} (2 u) + sin^{2} (u) = 0

Rewrite using trig identities

sin^{2} (2 u) + sin^{2} (u)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= (2 sin (u) cos (u))^{2} + sin^{2} (u)

Simplify

(2 sin (u) cos (u))^{2} + sin^{2} (u) : 4 sin^{2} (u) cos^{2} (u) + sin^{2} (u)

(2 sin (u) cos (u))^{2} + sin^{2} (u)

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} sin^{2} (u) cos^{2} (u) + sin^{2} (u)

2^{2} = 4

= 4 sin^{2} (u) cos^{2} (u) + sin^{2} (u)

= 4 sin^{2} (u) cos^{2} (u) + sin^{2} (u)

sin^{2} (u) + 4 cos^{2} (u) sin^{2} (u) = 0

Factor

sin^{2} (u) + 4 cos^{2} (u) sin^{2} (u) : sin^{2} (u) (4 cos^{2} (u) + 1)

sin^{2} (u) + 4 cos^{2} (u) sin^{2} (u)

Factor out common term

sin^{2} (u)

= sin^{2} (u) (1 + 4 cos^{2} (u))

sin^{2} (u) (4 cos^{2} (u) + 1) = 0

Solving each part separately

sin^{2} (u) = 0 or 4 cos^{2} (u) + 1 = 0

sin^{2} (u) = 0 : u = 2 πn, u = π + 2 πn

sin^{2} (u) = 0

Apply rule

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

sin (u) = 0

General solutions for

sin (u) = 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}

u = 0 + 2 πn, u = π + 2 πn

u = 0 + 2 πn, u = π + 2 πn

Solve

u = 0 + 2 πn : u = 2 πn

u = 0 + 2 πn

0 + 2 πn = 2 πn

u = 2 πn

u = 2 πn, u = π + 2 πn

4 cos^{2} (u) + 1 = 0 :

No Solution

4 cos^{2} (u) + 1 = 0

Solve by substitution

4 cos^{2} (u) + 1 = 0

Let:

cos (u) = u

4 u^{2} + 1 = 0

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

4 u^{2} + 1 = 0

Move

1

to the right side

4 u^{2} + 1 = 0

Subtract

1

from both sides

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

Simplify

4 u^{2} = - 1

4 u^{2} = - 1

Divide both sides by

4

4 u^{2} = - 1

Divide both sides by

4

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

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

u = - \frac{1}{4}, u = - - \frac{1}{4}

Simplify

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

Apply radical rule:

- a = - 1 a

- \frac{1}{4} = - 1 \frac{1}{4}

= - 1 \frac{1}{4}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= i \frac{1}{2}

Apply rule

1 = 1

= i \frac{1}{2}

Rewrite

i \frac{1}{2}

in standard complex form:

\frac{1}{2} i

i \frac{1}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 i}{2}

Multiply:

1 i = i

= \frac{i}{2}

= \frac{1}{2} i

Simplify

- - \frac{1}{4} : - i \frac{1}{2}

- - \frac{1}{4}

Simplify

- \frac{1}{4} : i \frac{1}{2}

- \frac{1}{4}

Apply radical rule:

- a = - 1 a

- \frac{1}{4} = - 1 \frac{1}{4}

= - 1 \frac{1}{4}

Apply imaginary number rule:

- 1 = i

= i \frac{1}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

\frac{1}{4} = \frac{1}{4}

= i \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= i \frac{1}{2}

= - i \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2} i

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

Substitute back

u = cos (u)

cos (u) = i \frac{1}{2}, cos (u) = - i \frac{1}{2}

cos (u) = i \frac{1}{2}, cos (u) = - i \frac{1}{2}

cos (u) = i \frac{1}{2} :

No Solution

cos (u) = i \frac{1}{2}

No Solution

cos (u) = - i \frac{1}{2} :

No Solution

cos (u) = - i \frac{1}{2}

No Solution

Combine all the solutions

No Solution

Combine all the solutions

u = 2 πn, u = π + 2 πn

Substitute back

u = 3 x

3 x = 2 πn : x = \frac{2 πn}{3}

3 x = 2 πn

Divide both sides by

3

3 x = 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{2 πn}{3}

Simplify

x = \frac{2 πn}{3}

x = \frac{2 πn}{3}

3 x = π + 2 πn : x = \frac{π + 2 πn}{3}

3 x = π + 2 πn

Divide both sides by

3

3 x = π + 2 πn

Divide both sides by

3

\frac{3 x}{3} = \frac{π}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} = \frac{π}{3} + \frac{2 πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{π}{3} + \frac{2 πn}{3} : \frac{π + 2 πn}{3}

\frac{π}{3} + \frac{2 πn}{3}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π + 2 πn}{3}

x = \frac{π + 2 πn}{3}

x = \frac{π + 2 πn}{3}

x = \frac{π + 2 πn}{3}

x = \frac{2 πn}{3}, x = \frac{π + 2 πn}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(6x)+sin^2(3x)=0 ?
The general solution for sin^2(6x)+sin^2(3x)=0 is x=(2pin)/3 ,x=(pi+2pin)/3