Hard Trig Equation?

Question

Solve a[sin(x)] + b[cos(y)] = c, on the interval
(0 < or = to theta < 2pi)

NOTE: a, b and c are constants and either a does not = 0 or b does not = 0.

Answer 1

I assume you mean

a sin(x) + b cos(x) = c, with 0<= x < 2pi .

Set y = sin(x); then cos(x) = sqrt(1 - sin^2(x)) = sqrt(1 - y^2)

a y + b sqrt(1 - y^2) = c
b sqrt(1-y^2) = c - ay
Square both sides;
b^2 (1-y^2) = c^2 -2ac y + a^y^2
y^2(a^2+b^2) - (2ac) y +(c^2-b^2) = 0

This is a quadratic in y; solve for y (2values), then set these = sin(x) and solve for x.

Answer 2

if you meant
a sin x + b cos x
you can write it as A cos(x-t)= A sin x sin t +A cos x cos t
so
A sin t = a
A cos t = b ans squaring and summing A^2 =a^2+b^2
so A = sqrt(a^2+b^2)
tan t = a/b so get t
The equation becomes
A cos(x-t)=c
so cos(x-t) = c/A