What is the general solution for sin(3x)=cos(2x) ?

The general solution for sin(3x)=cos(2x) is x=(pi+4pin)/(10),x=pi-pi/2+2pin

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sin(3x)=cos(2x)

Solution

sin (3 x) = cos (2 x)

Solution

x = \frac{π + 4 πn}{10}, x = π - \frac{π}{2} + 2 πn

Degrees

x = 1 8^{\circ} + 7 2^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (3 x) = cos (2 x)

Rewrite using trig identities

sin (3 x) = cos (2 x)

Use the following identity:

cos (x) = sin (\frac{π}{2} - x)

sin (3 x) = sin (\frac{π}{2} - 2 x)

sin (3 x) = sin (\frac{π}{2} - 2 x)

Apply trig inverse properties

sin (3 x) = sin (\frac{π}{2} - 2 x)

sin (x) = sin (y) \Rightarrow x = y + 2 π n, x = π - y + 2 π n

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

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

3 x = \frac{π}{2} - 2 x + 2 πn : x = \frac{π + 4 πn}{10}

3 x = \frac{π}{2} - 2 x + 2 πn

Move

2 x

to the left side

3 x = \frac{π}{2} - 2 x + 2 πn

Add

2 x

to both sides

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

Simplify

5 x = \frac{π}{2} + 2 πn

5 x = \frac{π}{2} + 2 πn

Divide both sides by

5

5 x = \frac{π}{2} + 2 πn

Divide both sides by

5

\frac{5 x}{5} = \frac{\frac{π}{2}}{5} + \frac{2 πn}{5}

Simplify

\frac{5 x}{5} = \frac{\frac{π}{2}}{5} + \frac{2 πn}{5}

Simplify

\frac{5 x}{5} : x

\frac{5 x}{5}

Divide the numbers:

\frac{5}{5} = 1

= x

Simplify

\frac{\frac{π}{2}}{5} + \frac{2 πn}{5} : \frac{π + 4 πn}{10}

\frac{\frac{π}{2}}{5} + \frac{2 πn}{5}

Apply rule

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

= \frac{\frac{π}{2} + 2 πn}{5}

Join

\frac{π}{2} + 2 πn : \frac{π + 4 πn}{2}

\frac{π}{2} + 2 πn

Convert element to fraction:

2 πn = \frac{2 πn 2}{2}

= \frac{π}{2} + \frac{2 πn \cdot 2}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{π + 2 πn \cdot 2}{2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π + 4 πn}{2}

= \frac{\frac{π + 4 πn}{2}}{5}

Apply the fraction rule:

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

= \frac{π + 4 πn}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{π + 4 πn}{10}

x = \frac{π + 4 πn}{10}

x = \frac{π + 4 πn}{10}

x = \frac{π + 4 πn}{10}

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

3 x = π - (\frac{π}{2} - 2 x) + 2 πn

Expand

π - (\frac{π}{2} - 2 x) + 2 πn : π - \frac{π}{2} + 2 x + 2 πn

π - (\frac{π}{2} - 2 x) + 2 πn

- (\frac{π}{2} - 2 x) : - \frac{π}{2} + 2 x

- (\frac{π}{2} - 2 x)

Distribute parentheses

= - (\frac{π}{2}) - (- 2 x)

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - \frac{π}{2} + 2 x

= π - \frac{π}{2} + 2 x + 2 πn

3 x = π - \frac{π}{2} + 2 x + 2 πn

Move

2 x

to the left side

3 x = π - \frac{π}{2} + 2 x + 2 πn

Subtract

2 x

from both sides

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

Simplify

x = π - \frac{π}{2} + 2 πn

x = π - \frac{π}{2} + 2 πn

x = \frac{π + 4 πn}{10}, x = π - \frac{π}{2} + 2 πn

x = \frac{π + 4 πn}{10}, x = π - \frac{π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(3x)=cos(2x) ?
The general solution for sin(3x)=cos(2x) is x=(pi+4pin)/(10),x=pi-pi/2+2pin