Finding Integer Values of k for Trigonometric Equation

We are given the equation $2\sin x = 2k + 1$, and we need to find the number of integer values of $k$ for which this equation has a solution for $x$.

For the equation to have a solution, the value of $2k + 1$ must be within the possible range of the expression $2\sin x$.

Understanding the Range of sin x

The sine function, $\sin x$, for any real number $x$, has a specific range of values. The range of $\sin x$ is from -1 to 1, inclusive.

Mathematically, this can be written as:

$$ -1 \le \sin x \le 1 $$

Determining the Range of 2sin x

Now, let's find the range of $2\sin x$. We can do this by multiplying the inequality for $\sin x$ by 2:

$$ 2 \times (-1) \le 2 \times \sin x \le 2 \times 1 $$

This gives us the range for $2\sin x$:

$$ -2 \le 2\sin x \le 2 $$

Setting up the Inequality for k

For the given equation $2\sin x = 2k + 1$ to have a solution, the value $2k + 1$ must fall within the range of $2\sin x$. Therefore, we must have:

$$ -2 \le 2k + 1 \le 2 $$

Solving the Inequality for k

To find the possible values of $k$, we need to isolate $k$ in the inequality. First, subtract 1 from all parts of the inequality:

$$ -2 - 1 \le (2k + 1) - 1 \le 2 - 1 $$

$$ -3 \le 2k \le 1 $$

Next, divide all parts of the inequality by 2:

$$ \frac{-3}{2} \le \frac{2k}{2} \le \frac{1}{2} $$

$$ -1.5 \le k \le 0.5 $$

Finding Integer Values of k

We are asked for the number of integer values of $k$ that satisfy the condition $-1.5 \le k \le 0.5$. The integers are whole numbers (positive, negative, or zero) with no fractional part.

Let's list the integers that are greater than or equal to -1.5 and less than or equal to 0.5:

• The first integer greater than or equal to -1.5 is -1.
• The next integer is 0.
• The next integer, 1, is greater than 0.5, so it is not included.

The integer values of $k$ that satisfy the inequality are -1 and 0.

Counting the Integer Values

The integer values of $k$ for which the equation $2\sin x = 2k + 1$ has a solution are -1 and 0. There are exactly two such integer values.

Thus, the number of integer values of $k$ is two.