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 1


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


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


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