The questions asks to find the intersections of
$$f(x) = 2 \sin(x-7) + 6$$ and $$g(x) = \cos(2x-10) + 8$$
within the interval $[6,14]$.
So my general strategy was, 1) equate the functions, 2) get all the $X$s on one side and 3) convert to the same trig function.
$$2 \sin(x-7) + 6 = \cos(2x-10) + 8$$
I recognized the double angle in the cosine function, so
$$2 \sin(x-7) + 6 = \cos[ 2 (x-5) ] + 8$$
$$2 \sin(x-7) + 6 = \cos^2(x-5) - \sin^2(x-5) + 8$$
$\cos^2$ can be replaced with an identity, so
$$2 \sin(x-7) + 6 = 1 - \sin^2(x-5) - \sin^2(x-5) + 8$$
Group like terms and move then around,
$$2 \sin(x-7) + 2 \sin^2(x-5) = 3$$
Extracting the $2$ from the left side.
$$\sin(x-7) + \sin^2(x-5) = \frac 3 2$$
So here is where I hit a mental wall.
I could use the sine addition formula, but that would reintroduce cosine.
I can't simplify the terms any further since the angles are different.
Where would I go from here? Or is my approach off completely?