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Chapter 2: Problem 8

Total no. of solutions of \([\sin x]+\cos x=0\); where \([\) ] denotes the greatest integer function, \(\cos x \in[0,2 \pi]\) is (a) 1 (b) 2 (c) 4 (d) 0

Short Answer

Expert verified
The correct answer is (a) 1, as there is only one solution for the given equation, which is \(x = \frac{\pi}{2}\).

Step by step solution

01

Rewrite the given equation

Firstly, let's rewrite the given equation as follows: \[[\sin x] + \cos x = 0\]
02

Analyze the range of sine function

Since the sine function varies between -1 and 1, it implies that the greatest integer function of sine varies between -1 and 0 (inclusive). Specifically: \[ [\sin x] = \begin{cases} -1, & \text{if } -1 \leq \sin x < 0 \\ 0, & \text{if } 0 \leq \sin x < 1 \\ \end{cases} \]
03

Separate the equations based on the range

Now, based on the range of sin(x), the given equation can be separated into two cases: Case 1: [-1 鈮� sin(x) < 0] \[-1 + \cos x = 0\] \[ \cos x = 1\] Case 2: [0 鈮� sin(x) < 1] \[0 + \cos x = 0\] \[ \cos x = 0\] Now, we'll solve these two cases separately and find the solutions for x between 0 and 2蟺.
04

Solve Case 1

For the equation: \(\cos x = 1\), the only solution for x in the interval [0, 2蟺] is: \[x = 0\] But, we must make sure this solution satisfies the original constraint [-1 鈮� sin(x) < 0] for case 1. Plugging x = 0: \[\sin(0) = 0\] In this case, sin(0) does not satisfy the constraint, so there is no solution in this case.
05

Solve Case 2

For the equation \(\cos x = 0\), there are two solutions for x in the interval [0, 2蟺]: \[x = \frac{\pi}{2}, \frac{3\pi}{2}\] Check if these solutions satisfy the original constraint [0 鈮� sin(x) < 1] for case 2: 1. sin(蟺/2) = 1, which satisfies the constraint. 2. sin(3蟺/2) = -1, which does not satisfy the constraint. So, there is only one valid solution for case 2: x = 蟺/2.
06

Count the total number of solutions and choose the correct answer

We found no solutions in case 1 and only one solution (x = 蟺/2) in case 2. This means that there is a total of only 1 solution. Therefore, the correct answer is (a) 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Greatest Integer Function
The greatest integer function, often represented as \([\lfloor x \rfloor]\), takes any real number and rounds it down to the nearest integer. This function is sometimes called the floor function.

Key points to remember include:
  • If \(x\) is already an integer, \(\lfloor x \rfloor = x\).
  • If \(x\) is not an integer, it moves to the next lowest integer.
In the context of trigonometric functions like sine, which can vary continuously between -1 and 1, the greatest integer function can take values -1 or 0, since:
  • The range \(-1 \leq \sin x < 0\) yields \([\sin x] = -1\).
  • The range \(0 \leq \sin x < 1\) yields \([\sin x] = 0\).
Understanding these conversions is essential when solving trigonometric equations involving the greatest integer function.
Trigonometric Functions
Trigonometric functions like sine (\(\sin\)) and cosine (\(\cos\)) are essential in mathematics for modeling periodic phenomena. These functions oscillate between specific values:
  • The sine function varies between -1 and 1.
  • The cosine function also varies between -1 and 1, with its cycle starting at 1 for \(x = 0\).
In solving equations, it's crucial to know these properties:
  • \(\sin(0) = 0\), reflecting that the sine graph crosses the x-axis at these points.
  • \(\cos(0) = 1\), meaning the cosine graph peaks at 1 initially.
For this exercise, ensuring that these values fall within the respective ranges specified by the greatest integer function is critical for identifying solutions.
Solution of Equations
Solving equations involves finding values of \(x\) that satisfy given mathematical conditions. Here, the task was to solve \([\sin x] + \cos x = 0\) with \(x\) ranging between 0 and \(2\pi\).
  • First, we consider the values of \([\sin x]\) through individual cases based on trigonometric properties.
  • Then \(\cos x\) must be algebraically extracted to examine possible solutions, ensuring they meet all initial criteria.
For example, in case 1:
  • Solving \(-1 + \cos x = 0\) gives \(\cos x = 1\), yet no valid \(x\) within \([-1 \leq \sin x < 0]\) satisfies the equation.
Case 2 involves solving \(0 + \cos x = 0\), yielding potential solutions of \([x = \frac{\pi}{2}, x = \frac{3\pi}{2}]\). However, \(\sin x\) must also fit within \([0 \leq \sin x < 1]\), thereby confirming \(x = \frac{\pi}{2}\) as the sole solution.
Understanding how to approach each case ensures clarity and accuracy when working through similar problems.

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Most popular questions from this chapter

Column-I (a) If \(\alpha, \beta\) are the solutions of \(\sin x=-\frac{\sqrt{3}}{2}\) in \([0,2 \pi]\) and \(\alpha, \gamma\) are the solutions of \(\cos x=-\frac{1}{2}\) in \([0,2 \pi]\), then (b) If \(\alpha, \beta\) are the solutions of \(\cot x=-\frac{1}{\sqrt{3}}\) in \([0,2 \pi]\) and \(\alpha, \beta\) are the solution of \(\operatorname{cosec} x=\frac{-2}{\sqrt{3}}\) in \([0,2 \pi]\), then (c) If \(\alpha, \beta\) are the solutions of \(\sin x=-\frac{\sqrt{3}}{2}\) in \([0,2 \pi]\) and \(\alpha, \gamma\) are the solutions of \(\tan x=\) \(\sqrt{3}\) in \([0,2 \pi]\), then Column-II (p) \(\alpha-\beta=\pi\) (q) \(\beta-\gamma=\pi\) (r) \(\alpha-\gamma=\pi\) (s) \(\alpha+\beta=3 \pi\) (t) \(\beta+\gamma=2 \pi\)

(b) \(1+\sin ^{5} x+\cos ^{5} x-3 / 2 \sin 2 x-0\) which can be written as \(1+(\sin x)^{3}+(\cos x)^{3}-3 \sin x \cdot \cos x=0\) Comparing this with \(\mathrm{a}^{3}+\mathrm{b}^{5}+\mathrm{c}^{3}-3 \mathrm{abc}=0\) $$ \begin{aligned} \therefore & a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\ &-(a+b+c)\left(\frac{1}{2}(a-b)^{2}+\frac{1}{2}(b-c)^{2}+\frac{1}{2}(c-a)^{2}\right)=0 \end{aligned}$$ only if \(\mathrm{a}+\mathrm{b}+\mathrm{c}=0\) or a \(-\mathrm{b}=\mathrm{c}\) \(\Rightarrow \sin x+\cos x+1=0 \Rightarrow \sin x+\cos x=-1\) \(\Rightarrow \quad \sqrt{2}\left(\cos \left(x-\frac{\pi}{4}\right)\right)=-1\) \(\Rightarrow \quad \cos \left(x-\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}\) \(\Rightarrow \quad \cos \left(x-\frac{\pi}{4}\right)=\cos \left(\frac{3 \pi}{4}\right)\) \(\Rightarrow \quad x-\frac{\pi}{4}=2 n \pi \pm \frac{3 \pi}{4}\) \(\Rightarrow \quad x=2 n \pi-\frac{\pi}{2}, 2 n \pi+\pi\) \(\Rightarrow \quad x-(2 n+1) \pi, 2 n \pi-\pi / 2 ; n \in \mathbb{Z}\)

Show that the equation \(\sin (x+\alpha)=a \sin 2 x+b\) has four roots whose sum is equal to \((2 n+1) \pi\) where \(n \in \mathbb{Z}\). Solution: \(\sin x \cos \alpha+\cos x \sin \alpha=2 a \sin x \cos x+b\) \(\Rightarrow \sin x \cos \alpha+\cos x \sin \alpha-2 a \sin x \cos x-b=0\)

Let \(\mathrm{f}(\mathrm{x})=\left(\mathrm{e}^{\mathrm{x}}-2\right)(\sin \mathrm{x}-\cos \mathrm{x})(\mathrm{x}-\ln 2)(\cos \mathrm{x}-1 / \sqrt{2})\) the roots are \(\mathrm{x}=\ln 2, \pi / 4,5 \pi / 4,7 \pi / 4,9 \pi / 4, \ldots\) \(\forall x \in(0, \ln 2)\), we get \(f(x)=(-)(-)(-)(+)<0\) \(\mathrm{x} \in(\ln 2, \pi / 4)\), we get \(\mathrm{f}(\mathrm{x})=(+)(-)(+)(+)<0\) \(\mathrm{x} \in(\pi / 4,5 \pi / 4)\), we get \(\mathrm{f}(\mathrm{x})=(+)(+)(+)(-)<0\) \(\Rightarrow\) Set of positive integer value for which \(f(x)<0\) where \(x\) \(\in(0, \pi)\) is \(\\{1,2,3\\}\) \(\Rightarrow x \in\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right) ; \mathrm{f}(\mathrm{x})=(+)(-)(+)(-)=\) positive The set \(\left(\frac{5 \pi}{4}, \frac{7 \pi}{4}\right)\) contains positive integers 4 and 5 . \(\Rightarrow\) Least positive integer for which \(f(x)>0\) is 4

The number of values of \(x\) in \([0,5 \pi]\) satisfying the equation \(3 \cos 2 x-10 \cos x+7=0\) is (a) 5 (b) 6 (c) 8 (d) 10
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