# how to solve this trigonometric equation involving one parameter?

I was trying to solve this question:

Let $$x$$ be defined from $$0$$ to $$\pi$$, in this equation $$\sin x$$ must be equal to "$$2-k$$". It has at least one solution if $$k$$ is in the range

• (1) from 1 to 2 (both included),
• (2) from 2 (included) to positive infinity,
• (3) from 1 to 3 (both included),
• (4) from negative infinity to 1 (included).

Here's my attempt to solve it:

Since x must be in the interval from 0 to pi,

if I plug 3 in the equation I'd get a negative value, so 2 and 3 are both incorrect.

the other options could be both correct,

but if I plug 0, I get "2-0 = 2", 2 it's not in the interval, and if I plug a negative value I got a value that is greater than 2, so it is still not correct.

The first option is correct for each value inside the interval. To prove that, I plug 1.5 (half) and it is acceptable (according to the equation), therefore my solution is (1).

is it correct? Let me know, and if you know a different method (not trial and error) let me know.

## 1 Answer

If $$x \in [0, \pi],$$ then $$0 \le \sin x \le 1.$$

Hence:

$$0 \le 2-k \le 1 \iff -2 \le -k \le -1 \iff 1 \le k \le 2.$$