SL 3.7—Circular functions, graphs, composites, transformations

Recall that $\sin (2x)=2\sin x\cos x$. Substituting gives $2(2\sin x\cos x)=\cos x-1$, i.e. $4\sin x\cos x-\cos x+1=0$.

Rearrange: $4\sin x\cos x-\cos x+1=0$. Factor $\cos x$: $\cos x(4\sin x-1)+1=0$. This does not factor cleanly, so use a graphical or numerical approach.

Graph $y=2\sin (2x)$ and $y=\cos x-1$ on $[0,2\pi )$. The function $\cos x-1$ ranges from $-2$ to $0$, while $2\sin (2x)$ oscillates between $-2$ and $2$. Counting intersections carefully (or using a GDC) reveals exactly 3 intersection points in the given domain.

The three solutions occur near $x\approx \pi$, $x\approx \frac{3\pi }{2}$, and one additional crossing. A GDC confirms 3 solutions in $[0,2\pi )$.

We need to count the number of solutions to $2\sin (2x)=\cos x-1$ in $[0,2\pi )$. The right side $\cos x-1\le 0$ always, so we are looking for where $2\sin (2x)$ equals a non-positive value.

2 solutions is too few — $2\sin (2x)$ completes two full cycles in $[0,2\pi )$ and intersects the curve $\cos x-1$ more than twice. Eliminate 2.

4 or 5 solutions would require more intersections than geometrically occur. Since $\cos x-1$ only reaches $0$ at $x=0$ and $x=2\pi$ (boundary), and touches $-2$ at $x=\pi$, the overlap regions are limited. A careful count rules out 4 and 5.

3 solutions is consistent with the graphical analysis. The answer is 3.

The function is $f(x)=3\sin \left(\frac{\pi }{6}(x-2)\right)+5$. Here $a=3$ (amplitude) and $d=5$ (vertical shift/midline).

Maximum value $=d+|a|=5+3=8$. Minimum value $=d-|a|=5-3=2$.

The maximum value is $8$ and the minimum value is $2$.

We need the maximum and minimum of the transformed sine function. The amplitude is $|a|=3$ and the midline is $y=5$.

This would require the function to have no vertical shift and amplitude 8. But $d=5\ne 0$ and $|a|=3\ne 8$. Eliminate this option.

This corresponds to amplitude 5 with no vertical shift, ignoring both $a=3$ and $d=5$. Incorrect.

This would imply amplitude $1$ centred on $y=5$. But amplitude is $3$, not $1$. Eliminate.

Maximum $=8$, Minimum $=2$ correctly uses $d+|a|=5+3=8$ and $d-|a|=5-3=2$.