What is the general solution for solvefor x,cos(qx)=sin(rx) ?

The general solution for solvefor x,cos(qx)=sin(rx) is x=(pi+4pin)/(2(r+q)),x=(pi+4pin)/(2(r-q))

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Popular >

solvefor x,cos(qx)=sin(rx)

Solution

solve for

x, cos (q x) = sin (r x)

Solution

x = \frac{π + 4 πn}{2 ( r + q )}, x = \frac{π + 4 πn}{2 ( r - q )}

Solution steps

cos (q x) = sin (r x)

Rewrite using trig identities

cos (q x) = sin (r x)

Use the following identity:

cos (x) = sin (\frac{π}{2} - x)

cos (q x) = sin (\frac{π}{2} - q x)

cos (q x) = sin (\frac{π}{2} - q x)

Apply trig inverse properties

cos (q x) = sin (\frac{π}{2} - q x)

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

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

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

r x = \frac{π}{2} - q x + 2 πn : x = \frac{π + 4 πn}{2 ( r + q )}; q \neq = - r

r x = \frac{π}{2} - q x + 2 πn

Move

q x

to the left side

r x = \frac{π}{2} - q x + 2 πn

Add

q x

to both sides

r x + q x = \frac{π}{2} - q x + 2 πn + q x

Simplify

r x + q x = \frac{π}{2} + 2 πn

r x + q x = \frac{π}{2} + 2 πn

Factor

r x + q x : x (r + q)

r x + q x

Factor out common term

x

= x (r + q)

x (r + q) = \frac{π}{2} + 2 πn

Divide both sides by

r + q; q \neq = - r

x (r + q) = \frac{π}{2} + 2 πn

Divide both sides by

r + q; q \neq = - r

\frac{x ( r + q )}{r + q} = \frac{\frac{π}{2}}{r + q} + \frac{2 πn}{r + q}; q \neq = - r

Simplify

\frac{x ( r + q )}{r + q} = \frac{\frac{π}{2}}{r + q} + \frac{2 πn}{r + q}

Simplify

\frac{x ( r + q )}{r + q} : x

\frac{x ( r + q )}{r + q}

Cancel the common factor:

r + q

= x

Simplify

\frac{\frac{π}{2}}{r + q} + \frac{2 πn}{r + q} : \frac{π + 4 πn}{2 ( r + q )}

\frac{\frac{π}{2}}{r + q} + \frac{2 πn}{r + q}

Apply rule

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

= \frac{\frac{π}{2} + 2 πn}{r + q}

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}}{r + q}

Apply the fraction rule:

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

= \frac{π + 4 πn}{2 ( r + q )}

x = \frac{π + 4 πn}{2 ( r + q )}; q \neq = - r

x = \frac{π + 4 πn}{2 ( r + q )}; q \neq = - r

x = \frac{π + 4 πn}{2 ( r + q )}; q \neq = - r

r x = π - (\frac{π}{2} - q x) + 2 πn : x = \frac{π + 4 πn}{2 ( r - q )}; q \neq = r

r x = π - (\frac{π}{2} - q x) + 2 πn

Expand

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

π - (\frac{π}{2} - q x) + 2 πn

= π - (\frac{π}{2} - x q) + 2 πn

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

- (\frac{π}{2} - q x)

Distribute parentheses

= - \frac{π}{2} - (- q x)

Apply minus-plus rules

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

= - \frac{π}{2} + q x

= π - \frac{π}{2} + q x + 2 πn

= π - \frac{π}{2} + x q + 2 πn

r x = π - \frac{π}{2} + x q + 2 πn

Move

x q

to the left side

r x = π - \frac{π}{2} + x q + 2 πn

Subtract

x q

from both sides

r x - x q = π - \frac{π}{2} + x q + 2 πn - x q

Simplify

r x - x q = π - \frac{π}{2} + 2 πn

r x - x q = π - \frac{π}{2} + 2 πn

Factor

r x - x q : x (r - q)

r x - x q

Factor out common term

x

= x (r - q)

x (r - q) = π - \frac{π}{2} + 2 πn

Divide both sides by

r - q; q \neq = r

x (r - q) = π - \frac{π}{2} + 2 πn

Divide both sides by

r - q; q \neq = r

\frac{x ( r - q )}{r - q} = \frac{π}{r - q} - \frac{\frac{π}{2}}{r - q} + \frac{2 πn}{r - q}; q \neq = r

Simplify

\frac{x ( r - q )}{r - q} = \frac{π}{r - q} - \frac{\frac{π}{2}}{r - q} + \frac{2 πn}{r - q}

Simplify

\frac{x ( r - q )}{r - q} : x

\frac{x ( r - q )}{r - q}

Cancel the common factor:

r - q

= x

Simplify

\frac{π}{r - q} - \frac{\frac{π}{2}}{r - q} + \frac{2 πn}{r - q} : \frac{π + 4 πn}{2 ( r - q )}

\frac{π}{r - q} - \frac{\frac{π}{2}}{r - q} + \frac{2 πn}{r - q}

Apply rule

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

= \frac{π - \frac{π}{2} + 2 πn}{r - q}

Join

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

π - \frac{π}{2} + 2 πn

Convert element to fraction:

π = \frac{π 2}{2}, 2 πn = \frac{2 πn 2}{2}

= \frac{π 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 - π + 2 πn \cdot 2}{2}

π 2 - π + 2 πn \cdot 2 = π + 4 πn

π 2 - π + 2 πn \cdot 2

Add similar elements:

2 π - π = π

= π + 2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= π + 4 πn

= \frac{π + 4 πn}{2}

= \frac{\frac{π + 4 πn}{2}}{r - q}

Apply the fraction rule:

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

= \frac{π + 4 πn}{2 ( r - q )}

x = \frac{π + 4 πn}{2 ( r - q )}; q \neq = r

x = \frac{π + 4 πn}{2 ( r - q )}; q \neq = r

x = \frac{π + 4 πn}{2 ( r - q )}; q \neq = r

x = \frac{π + 4 πn}{2 ( r + q )}, x = \frac{π + 4 πn}{2 ( r - q )}

x = \frac{π + 4 πn}{2 ( r + q )}, x = \frac{π + 4 πn}{2 ( r - q )}

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor x,cos(qx)=sin(rx) ?
The general solution for solvefor x,cos(qx)=sin(rx) is x=(pi+4pin)/(2(r+q)),x=(pi+4pin)/(2(r-q))