Solve 3sin(0.5x - 3) + 1 = 3 for all real x.

Step-by-step solution

1. 1

   Isolate the sine factor first. Nothing trigonometric can be done while the sine is buried under a multiplier and a shift. Subtract $1$, then divide by $3$:

   $3\sin(0.5x-3)=2\qquad\Longrightarrow\qquad \sin(0.5x-3)=\frac{2}{3}$

2. 2

   Confirm a solution exists. Sine only takes values in $[-1,1]$, and $\tfrac{2}{3}\approx0.667$ is inside that range, so solutions exist. (Had the right side exceeded $1$, the equation would have had none — this check is worth doing before any algebra.)

3. 3

   Write both branches of the general sine solution. Setting $\theta=0.5x-3$, the equation $\sin\theta=\tfrac23$ has two families per period, because sine takes each value once on the way up and once on the way down:

   $\theta=\arcsin\!\left(\tfrac{2}{3}\right)+2\pi k\qquad\text{or}\qquad \theta=\pi-\arcsin\!\left(\tfrac{2}{3}\right)+2\pi k,\quad k\in\mathbb{Z}$

   Keeping only the first branch is the single most common error here.

4. 4

   Undo the inner transformation. Add $3$ and multiply by $2$. Multiplying by $2$ also doubles the $2\pi k$ into $4\pi k$ — the period of $\sin(0.5x-3)$ in $x$ is $\tfrac{2\pi}{0.5}=4\pi$, not $2\pi$:

   $x=6+2\arcsin\!\left(\tfrac{2}{3}\right)+4\pi k\qquad\text{or}\qquad x=6+2\pi-2\arcsin\!\left(\tfrac{2}{3}\right)+4\pi k$

5. 5

   Evaluate the principal solutions numerically. $\arcsin\!\left(\tfrac23\right)\approx0.7297$, so the two base solutions are $x\approx6+1.4595=7.4595$ and $x\approx6+6.2832-1.4595=10.8237$.

6. 6

   Verify both by substitution. For $x\approx7.4595$: $0.5x-3\approx0.7297$ and $3\sin(0.7297)+1=3.0000$ ✓. For $x\approx10.8237$: $0.5x-3\approx2.4119$ and $3\sin(2.4119)+1=3.0000$ ✓. Adding any multiple of $4\pi$ to either reproduces the same value.