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2sin(x)sin(3x)=1,0<= x<= 2pi

Solution

2 sin (x) sin (3 x) = 1, 0 \leq x \leq 2 π

Solution

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

Degrees

x = 3 0^{\circ}, x = 15 0^{\circ}, x = 21 0^{\circ}, x = 33 0^{\circ}, x = 4 5^{\circ}, x = 13 5^{\circ}, x = 22 5^{\circ}, x = 31 5^{\circ}

Solution steps

2 sin (x) sin (3 x) = 1, 0 \leq x \leq 2 π

Subtract

1

from both sides

2 sin (x) sin (3 x) - 1 = 0

Rewrite using trig identities

- 1 + 2 sin (3 x) sin (x)

sin (3 x) = 3 sin (x) - 4 sin^{3} (x)

sin (3 x)

Rewrite using trig identities

sin (3 x)

Rewrite as

= sin (2 x + x)

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

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

Use the Double Angle identity:

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

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

Simplify

cos (2 x) sin (x) + cos (x) \cdot 2 sin (x) cos (x) : sin (x) cos (2 x) + 2 cos^{2} (x) sin (x)

cos (2 x) sin (x) + cos (x) 2 sin (x) cos (x)

cos (x) \cdot 2 sin (x) cos (x) = 2 cos^{2} (x) sin (x)

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

Apply exponent rule:

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

cos (x) cos (x) = cos^{1 + 1} (x)

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

Add the numbers:

1 + 1 = 2

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

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

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

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

Use the Double Angle identity:

cos (2 x) = 1 - 2 sin^{2} (x)

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

Use the Pythagorean identity:

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

cos^{2} (x) = 1 - sin^{2} (x)

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

Expand

(1 - 2 sin^{2} (x)) sin (x) + 2 (1 - sin^{2} (x)) sin (x) : - 4 sin^{3} (x) + 3 sin (x)

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

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

Expand

sin (x) (1 - 2 sin^{2} (x)) : sin (x) - 2 sin^{3} (x)

sin (x) (1 - 2 sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = sin (x), b = 1, c = 2 sin^{2} (x)

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

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

Simplify

1 \cdot sin (x) - 2 sin^{2} (x) sin (x) : sin (x) - 2 sin^{3} (x)

1 sin (x) - 2 sin^{2} (x) sin (x)

1 \cdot sin (x) = sin (x)

1 sin (x)

Multiply:

1 \cdot sin (x) = sin (x)

= sin (x)

2 sin^{2} (x) sin (x) = 2 sin^{3} (x)

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

Apply exponent rule:

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

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

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

Add the numbers:

2 + 1 = 3

= 2 sin^{3} (x)

= sin (x) - 2 sin^{3} (x)

= sin (x) - 2 sin^{3} (x)

= sin (x) - 2 sin^{3} (x) + 2 (1 - sin^{2} (x)) sin (x)

Expand

2 sin (x) (1 - sin^{2} (x)) : 2 sin (x) - 2 sin^{3} (x)

2 sin (x) (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 2 sin (x), b = 1, c = sin^{2} (x)

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

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

Simplify

2 \cdot 1 \cdot sin (x) - 2 sin^{2} (x) sin (x) : 2 sin (x) - 2 sin^{3} (x)

2 \cdot 1 sin (x) - 2 sin^{2} (x) sin (x)

2 \cdot 1 \cdot sin (x) = 2 sin (x)

2 \cdot 1 sin (x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 sin (x)

2 sin^{2} (x) sin (x) = 2 sin^{3} (x)

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

Apply exponent rule:

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

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

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

Add the numbers:

2 + 1 = 3

= 2 sin^{3} (x)

= 2 sin (x) - 2 sin^{3} (x)

= 2 sin (x) - 2 sin^{3} (x)

= sin (x) - 2 sin^{3} (x) + 2 sin (x) - 2 sin^{3} (x)

Simplify

sin (x) - 2 sin^{3} (x) + 2 sin (x) - 2 sin^{3} (x) : - 4 sin^{3} (x) + 3 sin (x)

sin (x) - 2 sin^{3} (x) + 2 sin (x) - 2 sin^{3} (x)

Group like terms

= - 2 sin^{3} (x) - 2 sin^{3} (x) + sin (x) + 2 sin (x)

Add similar elements:

- 2 sin^{3} (x) - 2 sin^{3} (x) = - 4 sin^{3} (x)

= - 4 sin^{3} (x) + sin (x) + 2 sin (x)

Add similar elements:

sin (x) + 2 sin (x) = 3 sin (x)

= - 4 sin^{3} (x) + 3 sin (x)

= - 4 sin^{3} (x) + 3 sin (x)

= - 4 sin^{3} (x) + 3 sin (x)

= - 1 + 2 (3 sin (x) - 4 sin^{3} (x)) sin (x)

- 1 + (3 sin (x) - 4 sin^{3} (x)) \cdot 2 sin (x) = 0

Solve by substitution

- 1 + (3 sin (x) - 4 sin^{3} (x)) \cdot 2 sin (x) = 0

Let:

sin (x) = u

- 1 + (3 u - 4 u^{3}) \cdot 2 u = 0

- 1 + (3 u - 4 u^{3}) \cdot 2 u = 0 : u = \frac{1}{2}, u = - \frac{1}{2}, u = \frac{1}{2}, u = - \frac{1}{2}

- 1 + (3 u - 4 u^{3}) \cdot 2 u = 0

Expand

- 1 + (3 u - 4 u^{3}) \cdot 2 u : - 1 + 6 u^{2} - 8 u^{4}

- 1 + (3 u - 4 u^{3}) \cdot 2 u

= - 1 + 2 u (3 u - 4 u^{3})

Expand

2 u (3 u - 4 u^{3}) : 6 u^{2} - 8 u^{4}

2 u (3 u - 4 u^{3})

Apply the distributive law:

a (b - c) = ab - a c

a = 2 u, b = 3 u, c = 4 u^{3}

= 2 u \cdot 3 u - 2 u \cdot 4 u^{3}

= 2 \cdot 3 uu - 2 \cdot 4 u^{3} u

Simplify

2 \cdot 3 uu - 2 \cdot 4 u^{3} u : 6 u^{2} - 8 u^{4}

2 \cdot 3 uu - 2 \cdot 4 u^{3} u

2 \cdot 3 uu = 6 u^{2}

2 \cdot 3 uu

Multiply the numbers:

2 \cdot 3 = 6

= 6 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 6 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 6 u^{2}

2 \cdot 4 u^{3} u = 8 u^{4}

2 \cdot 4 u^{3} u

Multiply the numbers:

2 \cdot 4 = 8

= 8 u^{3} u

Apply exponent rule:

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

u^{3} u = u^{3 + 1}

= 8 u^{3 + 1}

Add the numbers:

3 + 1 = 4

= 8 u^{4}

= 6 u^{2} - 8 u^{4}

= 6 u^{2} - 8 u^{4}

= - 1 + 6 u^{2} - 8 u^{4}

- 1 + 6 u^{2} - 8 u^{4} = 0

Write in the standard form

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

- 8 u^{4} + 6 u^{2} - 1 = 0

Rewrite the equation with

v = u^{2}

and

v^{2} = u^{4}

- 8 v^{2} + 6 v - 1 = 0

Solve

- 8 v^{2} + 6 v - 1 = 0 : v = \frac{1}{4}, v = \frac{1}{2}

- 8 v^{2} + 6 v - 1 = 0

Solve with the quadratic formula

- 8 v^{2} + 6 v - 1 = 0

Quadratic Equation Formula:

For

a = - 8, b = 6, c = - 1

v_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 8 ) ( - 1 )}{2 ( - 8 )}

v_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 8 ) ( - 1 )}{2 ( - 8 )}

6^{2} - 4 (- 8) (- 1) = 2

6^{2} - 4 (- 8) (- 1)

Apply rule

- (- a) = a

= 6^{2} - 4 \cdot 8 \cdot 1

Multiply the numbers:

4 \cdot 8 \cdot 1 = 32

= 6^{2} - 32

6^{2} = 36

= 36 - 32

Subtract the numbers:

36 - 32 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

v_{1, 2} = \frac{- 6 \pm 2}{2 ( - 8 )}

Separate the solutions

v_{1} = \frac{- 6 + 2}{2 ( - 8 )}, v_{2} = \frac{- 6 - 2}{2 ( - 8 )}

v = \frac{- 6 + 2}{2 ( - 8 )} : \frac{1}{4}

\frac{- 6 + 2}{2 ( - 8 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 + 2}{- 2 \cdot 8}

Add/Subtract the numbers:

- 6 + 2 = - 4

= \frac{- 4}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 4}{- 16}

Apply the fraction rule:

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

= \frac{4}{16}

Cancel the common factor:

4

= \frac{1}{4}

v = \frac{- 6 - 2}{2 ( - 8 )} : \frac{1}{2}

\frac{- 6 - 2}{2 ( - 8 )}

Remove parentheses:

(- a) = - a

= \frac{- 6 - 2}{- 2 \cdot 8}

Subtract the numbers:

- 6 - 2 = - 8

= \frac{- 8}{- 2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{- 8}{- 16}

Apply the fraction rule:

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

= \frac{8}{16}

Cancel the common factor:

8

= \frac{1}{2}

The solutions to the quadratic equation are:

v = \frac{1}{4}, v = \frac{1}{2}

v = \frac{1}{4}, v = \frac{1}{2}

Substitute back

v = u^{2},

solve for

u

Solve

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

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}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

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

- \frac{1}{4}

Simplify

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

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2}

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

Solve

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

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

For

x^{2} = f (a)

the solutions are

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

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

The solutions are

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

Substitute back

u = sin (x)

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

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

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

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

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

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

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{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{4}, x = \frac{3 π}{4}

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

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}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{4}, x = \frac{3 π}{4}

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{5 π}{4}, x = \frac{7 π}{4}

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

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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{5 π}{4}, x = \frac{7 π}{4}

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}, x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = \frac{π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}, x = \frac{7 π}{4}

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