# 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?

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Is it $x-7$ or $x+7$? – Amzoti Dec 22 '13 at 22:09
Maybe consider plotting, including a parametric plot, see both using WA. Look at the point $6$ in the parametric plot, which is a hint. – Amzoti Dec 22 '13 at 22:22
Excuse my typos. I corrected it to x-7. – Marty B. Dec 22 '13 at 22:23
I graphed the functions on Desmos and I can see that there are two solutions in the given interval. However, I still don't know how to derive the answer. – Marty B. Dec 22 '13 at 22:31

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