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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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Most popular questions from this chapter

Let two numbers have arithmetic mean 9 and geometric mean 4 then these numbers are the roots of the quadratic equation (a) \(x^{2}+18 x+16=0\) (b) \(x^{2}-18 x+16=0\) (c) \(x^{2}+18 x-16=0\) (d) \(x^{2}-18 x-16=0\)

The number of real values of \(\mathrm{x}\) satisfying the equation \(3\left[x^{2}+\left(1 / x^{2}\right)\right]-16[x+(1 / x)]+26=0\) is (a) 1 (b) 2 (c) 3 (d) 4

If the roots of the quadratic equation \((2 \mathrm{k}+3) \mathrm{x}^{2}+2(\mathrm{k}+3)\) \(x+(k+5)=0\\{k \in R, k \neq\\{(-3) / 2\\}\\}\) are equal, then \(\mathrm{K}=\) (a) 1,6 (b) \(-1,-6\) (c) \(-1,6\) (d) \(1,-6\)

For the equation \(\ell \mathrm{x}^{2}+\mathrm{mx}+\mathrm{n} \pm 0, \ell \neq 0\) If \(\alpha \& \beta\) are roots of equation and \(m^{3}+\ell^{2} n+\ell n^{2}=3 \ell m n\) then (a) \(\alpha=\beta^{2}\) (b) \(\alpha^{3}=\beta\) (c) \(\alpha+\beta=\alpha \beta\) (d) \(\alpha \beta=1\)

If the product of roots of equation \(x^{2}-5 k x+4 e^{4 \log k}-3=0\) is 61 then \(\mathrm{k}\) is \(\ldots \ldots \ldots \ldots\) (a) 1 (b) 2 (c) 3 (d) 4
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