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Popular >

sin(3x)+cos(2x)=0

Solution

sin (3 x) + cos (2 x) = 0

Solution

x = \frac{3 π}{2} + 2 πn, x = 0.94247 \dots + 2 πn, x = π - 0.94247 \dots + 2 πn, x = - 0.31415 \dots + 2 πn, x = π + 0.31415 \dots + 2 πn

Degrees

x = 27 0^{\circ} + 36 0^{\circ} n, x = 5 4^{\circ} + 36 0^{\circ} n, x = 12 6^{\circ} + 36 0^{\circ} n, x = - 1 8^{\circ} + 36 0^{\circ} n, x = 19 8^{\circ} + 36 0^{\circ} n

Solution steps

sin (3 x) + cos (2 x) = 0

Rewrite using trig identities

cos (2 x) + sin (3 x)

Use the Double Angle identity:

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

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

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

Solve by substitution

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

Let:

sin (x) = u

1 - 2 u^{2} + 3 u - 4 u^{3} = 0

1 - 2 u^{2} + 3 u - 4 u^{3} = 0 : u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

1 - 2 u^{2} + 3 u - 4 u^{3} = 0

Write in the standard form

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

- 4 u^{3} - 2 u^{2} + 3 u + 1 = 0

Factor

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

- 4 u^{3} - 2 u^{2} + 3 u + 1

Factor out common term

- 1

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

Factor

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

4 u^{3} + 2 u^{2} - 3 u - 1

Use the rational root theorem

a_{0} = 1, a_{n} = 4

The dividers of

a_{0} : 1,

The dividers of

a_{n} : 1, 2, 4

Therefore, check the following rational numbers:

\pm \frac{1}{1 , 2 , 4}

- \frac{1}{1}

is a root of the expression, so factor out

u + 1

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

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} = 4 u^{2} - 2 u - 1

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1}

Divide

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} : \frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} = 4 u^{2} + \frac{- 2 u ^{2} - 3 u - 1}{u + 1}

Divide the leading coefficients of the numerator

4 u^{3} + 2 u^{2} - 3 u - 1

and the divisor

u + 1 : \frac{4 u ^{3}}{u} = 4 u^{2}

Quotient = 4 u^{2}

Multiply

u + 1

4 u^{2} : 4 u^{3} + 4 u^{2}

Subtract

4 u^{3} + 4 u^{2}

from

4 u^{3} + 2 u^{2} - 3 u - 1

to get new remainder

Remainder = - 2 u^{2} - 3 u - 1

Therefore

\frac{4 u ^{3} + 2 u ^{2} - 3 u - 1}{u + 1} = 4 u^{2} + \frac{- 2 u ^{2} - 3 u - 1}{u + 1}

= 4 u^{2} + \frac{- 2 u ^{2} - 3 u - 1}{u + 1}

Divide

\frac{- 2 u ^{2} - 3 u - 1}{u + 1} : \frac{- 2 u ^{2} - 3 u - 1}{u + 1} = - 2 u + \frac{- u - 1}{u + 1}

Divide the leading coefficients of the numerator

- 2 u^{2} - 3 u - 1

and the divisor

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

Quotient = - 2 u

Multiply

u + 1

- 2 u : - 2 u^{2} - 2 u

Subtract

- 2 u^{2} - 2 u

from

- 2 u^{2} - 3 u - 1

to get new remainder

Remainder = - u - 1

Therefore

\frac{- 2 u ^{2} - 3 u - 1}{u + 1} = - 2 u + \frac{- u - 1}{u + 1}

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

Divide

\frac{- u - 1}{u + 1} : \frac{- u - 1}{u + 1} = - 1

Divide the leading coefficients of the numerator

- u - 1

and the divisor

u + 1 : \frac{- u}{u} = - 1

Quotient = - 1

Multiply

u + 1

- 1 : - u - 1

Subtract

- u - 1

from

- u - 1

to get new remainder

Remainder = 0

Therefore

\frac{- u - 1}{u + 1} = - 1

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

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

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

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Solve

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = 4, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 \cdot 4 (- 1) = 25

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

Apply rule

- (- a) = a

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

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

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

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

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 2^{2} + 16

2^{2} = 4

= 4 + 16

Add the numbers:

4 + 16 = 20

= 20

Prime factorization of

20 : 2^{2} \cdot 5

20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 5

= 2^{2} \cdot 5

= 2^{2} \cdot 5

Apply radical rule:

= 5 2^{2}

Apply radical rule:

2^{2} = 2

= 25

u_{1, 2} = \frac{- ( - 2 ) \pm 2 5}{2 \cdot 4}

Separate the solutions

u_{1} = \frac{- ( - 2 ) + 2 5}{2 \cdot 4}, u_{2} = \frac{- ( - 2 ) - 2 5}{2 \cdot 4}

u = \frac{- ( - 2 ) + 2 5}{2 \cdot 4} : \frac{1 + 5}{4}

\frac{- ( - 2 ) + 2 5}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{2 + 2 5}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 + 2 5}{8}

Factor

2 + 25 : 2 (1 + 5)

2 + 25

Rewrite as

= 2 \cdot 1 + 25

Factor out common term

2

= 2 (1 + 5)

= \frac{2 ( 1 + 5 )}{8}

Cancel the common factor:

2

= \frac{1 + 5}{4}

u = \frac{- ( - 2 ) - 2 5}{2 \cdot 4} : \frac{1 - 5}{4}

\frac{- ( - 2 ) - 2 5}{2 \cdot 4}

Apply rule

- (- a) = a

= \frac{2 - 2 5}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{2 - 2 5}{8}

Factor

2 - 25 : 2 (1 - 5)

2 - 25

Rewrite as

= 2 \cdot 1 - 25

Factor out common term

2

= 2 (1 - 5)

= \frac{2 ( 1 - 5 )}{8}

Cancel the common factor:

2

= \frac{1 - 5}{4}

The solutions to the quadratic equation are:

u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

The solutions are

u = - 1, u = \frac{1 + 5}{4}, u = \frac{1 - 5}{4}

Substitute back

u = sin (x)

sin (x) = - 1, sin (x) = \frac{1 + 5}{4}, sin (x) = \frac{1 - 5}{4}

sin (x) = - 1, sin (x) = \frac{1 + 5}{4}, sin (x) = \frac{1 - 5}{4}

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

sin (x) = - 1

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

sin (x) = \frac{1 + 5}{4} : x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

sin (x) = \frac{1 + 5}{4}

Apply trig inverse properties

sin (x) = \frac{1 + 5}{4}

General solutions for

sin (x) = \frac{1 + 5}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn

sin (x) = \frac{1 - 5}{4} : x = arcsin (\frac{1 - 5}{4}) + 2 πn, x = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

sin (x) = \frac{1 - 5}{4}

Apply trig inverse properties

sin (x) = \frac{1 - 5}{4}

General solutions for

sin (x) = \frac{1 - 5}{4}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{1 - 5}{4}) + 2 πn, x = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

x = arcsin (\frac{1 - 5}{4}) + 2 πn, x = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

Combine all the solutions

x = \frac{3 π}{2} + 2 πn, x = arcsin (\frac{1 + 5}{4}) + 2 πn, x = π - arcsin (\frac{1 + 5}{4}) + 2 πn, x = arcsin (\frac{1 - 5}{4}) + 2 πn, x = π + arcsin (- \frac{1 - 5}{4}) + 2 πn

Show solutions in decimal form

x = \frac{3 π}{2} + 2 πn, x = 0.94247 \dots + 2 πn, x = π - 0.94247 \dots + 2 πn, x = - 0.31415 \dots + 2 πn, x = π + 0.31415 \dots + 2 πn

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