Intersection of Trig Functions 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.
So
$$2 \sin(x-7) + 6 = \cos(2x-10) + 8$$
I recognized the double angle in the cosine function, so
$$2 \sin(x-7) + 6 = \cos[ 2 (x-5) ] + 8$$
then
$$2 \sin(x-7) + 6 = \cos^2(x-5) - \sin^2(x-5) + 8$$
$\cos^2$ 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?

$$\sin(x-7)+\sin^2(x-5)=\frac32\iff\sin(t-2)+\sin^2t=\frac32$$
$$\sin t\cdot\cos(-2)+\cos t\cdot\sin(-2)+\sin^2t=\frac32$$
$$\sin^2t+\cos2\cdot\sin t-\sin2\cdot\cos t=\frac32\iff y^2+\cos2\cdot y-\sin2\cdot\bigg(\!\!\pm\sqrt{1-y^2}\bigg)=\frac32$$
$$y^2+\cos2\cdot y-\frac32=\sin2\cdot\bigg(\!\!\pm\sqrt{1-y^2}\bigg)\iff\bigg(y^2+\cos2\cdot y-\frac32\bigg)^2=\sin^22\cdot(1-y^2)$$
Now you're left with solving a quartic equation in $y=\sin t=\sin(x-5)\iff x=5+\arcsin y$
