Solving Trig Equations with Calculators, Part II — Question 5

Question 5 - Solution

Step 1: Isolate the trigonometric function. The equation becomes

$$\sin(1x)=0.65,\qquad u=1x,\qquad 0\le u\le 2\pi.$$

Let $\alpha=\arcsin(0.65)$. All solutions are

$$u=\alpha+2k\pi\quad\text{or}\quad u=\pi-\alpha+2k\pi,\qquad k\in\mathbb Z.$$

Step 2: Restrict and convert. Keep precisely the values of $u$ in $[0,2\pi]$ and divide by $1$.

Evaluating the inverse function at full precision and rounding only the final values gives

$$\boxed{x\approx 0.71,\ 2.43}.$$

These are all 2 solutions in the stated interval, in radians. Substitution of the unrounded values verifies the original equation.