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

solve for

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

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 (\frac{π}{2} - q x)

Apply trig inverse properties

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

sin (21 6^{\circ}) = sin (θ)

cos (θ) = - 0.2

sin^{2} (x) - 5 cos (x) = 3

cos (x) = - \frac{9}{41}

5 sin (y) = 3 sin (y) + 3

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))