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Chapter 3: Problem 213

The number of values of \(\mathrm{x}\) in the interval \([0,3 \pi]\) Satisfying the equation \(2 \sin ^{2} \mathrm{x}+5 \sin \mathrm{x}-3=0\) is (a) 6 (b) 1 (c) 2 (d) 4

Short Answer

Expert verified
The number of values of \(x\) in the interval \([0, 3\pi]\) satisfying the equation \(2\sin^2{x}+5\sin{x}-3=0\) is 2. Hence, the correct answer is (c) 2.

Step by step solution

01

Identify the equation type

The given equation is quadratic in nature as it contains \(\sin^2 x\). So we can rewrite it as a quadratic equation with \(y = \sin x\), so it becomes \(2y^2 + 5y - 3 = 0\).
02

Find the roots of the quadratic equation

We'll use the quadratic formula to find the roots of the equation \(2y^2 + 5y - 3 = 0\). The quadratic formula is given by \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where a, b, and c are the coefficients of the quadratic equation \(ay^2 + by + c = 0\). In our case, \(a = 2\), \(b = 5\), and \(c = -3\). So we have: \[y = \frac{-(5) \pm \sqrt{(5)^2 - 4(2)(-3)}}{2(2)}\] Solving for y, we get two roots: \(y_1 = \frac{1}{2}, y_2 = -3\)
03

Convert the roots back to x

Now we have the roots y as \(y_1 = \frac{1}{2}\) and \(y_2 = -3\). Next, we need to convert them back to x values. \(y = \sin x\) implies that \(x = \arcsin y\). The general solutions of \(x\) for these two roots are: \(x_1 = \arcsin \frac{1}{2} + 2n\pi, x_2 = \arcsin (-3) + 2m\pi\), where n and m are integers. However, since the sine function has a range of \([-1, 1]\), the second equation has no real values for x as \(\arcsin (-3)\) is undefined. Thus, we can disregard it.
04

Determine the solutions in the given interval

We now need to find the values of x in the interval \([0, 3 \pi]\) that satisfy the equation. Using \(x_1 = \arcsin \frac{1}{2} + 2n\pi\), we'll find the possible values of x. For \(n = 0\): \(x_1 = \arcsin \frac{1}{2} + 2(0)\pi = \frac{\pi}{6}\) For \(n = 1\): \(x_1 = \arcsin \frac{1}{2} + 2(1)\pi = \frac{\pi}{6} + 2\pi\) For \(n = 2\): \(x_1 = \arcsin \frac{1}{2} + 2(2)\pi = \frac{\pi}{6} + 4\pi\) Since \(0 \le x \le 3\pi\), the only two values of x that lie within this interval are \(\frac{\pi}{6}\) and \(\frac{\pi}{6} + 2\pi\). So, the number of values of x in the interval \([0, 3 \pi]\) that satisfy the given equation is 2. Hence, the correct answer is (c) 2.

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