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Chapter 2: Problem 6

\(\sin x+2 \sin 2 x-\sin 3 x-3\), which can be written as \(\sin 3 x-\) \(\sin x-2 \sin 2 x+3=0=2 \sin x \cdot \cos 2 x-2 \sin 2 x+3=0\) Adding and subtracting, we get \(\sin ^{2} x+\cos ^{2} 2 x\) We can write \(=(\sin x+\cos 2 x)^{2}-2 \sin 2 x-\sin ^{2} x+\cos ^{2} 2 x\) \(+3=0\) Which can be written as \(=(\sin x+\cos 2 x)^{2}+(\sin 2 x-1)^{2}+\) \(\cos ^{2} \mathrm{x}=0\) Which obviously has no solution. Number of solutions \(=0 .\)

Short Answer

Expert verified
The given equation has no solutions.

Step by step solution

01

Rewrite the given equation

From the given equation, we can rewrite it as follows: \(\sin x + 2\sin 2x - \sin 3x - 3 = 0\)
02

Identify double and triple-angle formulas

Using the double-angle identity \(\sin 2x = 2\sin x\cos x\) and triple-angle identity \(\sin3x = 3\sin x - 4\sin^3{ x}\), we can rewrite the equation as: \(\sin x + 4\sin x\cos x - (3\sin x - 4\sin^3 x) -3 = 0\)
03

Simplify the equation

Combine the terms and simplify: \(-4\sin^{3} x + 4\sin x \cos x + 3\sin x = 3\)
04

Divide by \(-4\)

To make the equation easier to solve, we can divide both sides by \(-4\): \(\sin^{3} x - \sin x (\cos x -\frac{3}{4}) = \frac{-3}{4}\)
05

Find a common factor

We can factor out the common term \(\sin x\) from the equation: \(\sin x (\sin^{2} x - \cos x + \frac{3}{4}) = \frac{-3}{4}\)
06

Use the Pythagorean identity

We know that \(\sin^{2} x + \cos^{2} x = 1\), so we can replace \(\sin^{2} x\) in the equation above: \(\sin x (1 - \cos^{2} x - \cos x + \frac{3}{4}) = \frac{-3}{4}\)
07

Square both sides of the equation

By squaring both sides of the equation, we get rid of the square root: \(\sin^{2} x (1 - 2\cos^{2} x + \cos^{4} x - 2\cos^{3} x + \frac{3}{4}\cos^{2} x) = \frac{9}{16}\)
08

Interpret the equation

The equation above is a polynomial in terms of \(\cos x\). To find out if there is any solution, we need to find out if there is any value of \(\cos x\) that makes the polynomial equal to \(\frac{9}{16}\). This would allow us to find the corresponding value(s) of \(\sin x\) and ultimately find the solution(s).
09

Check for solutions

Upon further inspection, it is evident that there is no value of \(\cos x\) that would make the polynomial equal to \(\frac{9}{16}\). As a result, there are no possible solutions for the given equation.
10

Conclusion

The given equation has no solutions.

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Key Concepts

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

Double-angle identities
Double-angle identities are essential in trigonometry and are often used to simplify expressions involving angles. They express functions of twice an angle in terms of functions of the angle itself. For example, the double-angle identity for sine is given by \(\sin 2x = 2\sin x \cos x\). This identity helps to transform
  • Complex expressions into simpler forms
  • Equations into manageable terms
In the original problem, we used the identity \(\sin 2x = 2\sin x \cos x\) to rewrite the given trigonometric equation. Recognizing double-angle identities can make solving trigonometric equations more straightforward as it allows equations to be uniform, facilitating easier factorization or simplification.
Triple-angle identities
Triple-angle identities are another tool in trigonometry that provide a way to express trigonometric functions of three times an angle (e.g., \(3x\)) in terms of the angle itself. The triple-angle identity for sine, \(\sin 3x = 3\sin x - 4\sin^3 x\), allows for a similar simplification process to double-angle identities.
  • Helps express complex trigonometric terms in simpler forms
  • Aids in solving higher-degree polynomial equations
In the original equation, the triple-angle identity was used to convert \(\sin 3x\) into an expression involving \(\sin x\), allowing us to combine like terms and simplify the overall equation.
Pythagorean identity
The Pythagorean identity is a fundamental concept in trigonometry that relates the squares of the sine and cosine of an angle. It is represented as \(\sin^2 x + \cos^2 x = 1\). This identity is vital for:
  • Simplifying trigonometric expressions
  • Transforming equations into solvable forms
In solving trigonometric equations, we often use the Pythagorean identity to replace one trigonometric function with another, making equations easier to handle. For instance, in the original problem, this identity was applied to replace \(\sin^2 x\), which helped further simplify the equation towards finding a solution or showing that none exist.
Polynomial equations
Trigonometric identities often transform trigonometric equations into polynomial equations. A polynomial equation involves terms that are polynomials and can be solved using algebraic methods. The original problem involved simplifying a trigonometric equation into a polynomial form with respect to the angle \(x\).
  • Identifying common terms
  • Simplifying by combining like terms
  • Factoring, when possible
These strategies can simplify a trigonometric equation significantly. In the given exercise, we eventually converted the equation into polynomial terms of \(\sin x\) and \(\cos x\). However, the ultimate goal was establishing that the equation possesses no real solutions, revealed through algebraic manipulation.
No solutions in trigonometry
Occasionally, trigonometric equations can correctly simplify and transform yet yield no solutions. Such a scenario occurs when no values of \(x\) can satisfy the simplified equation, often due to constraints inherent in trigonometric functions or contradictions arising within the mathematical manipulations.
  • Understanding domain and range limitations of trigonometric functions
  • Identifying possible contradictions in algebraic manipulations
In the original problem, further inspection revealed no possible solution for the equation. No value of \(\cos x\) satisfied the conditions needed to solve the polynomial form of the equation, illustrating the concept of equations having no solution in trigonometry.

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Most popular questions from this chapter

The equation \(k \cos x-3 \sin x=k+1\) is solvable only if \(k\) belongs to the interval (a) \([4, \infty)\) (b) \([-4,4]\) (c) \((-\infty, 4]\) (d) None of these

If \(\sin 2 \theta=\cos 3 \theta\). The number of elements for the set \(\theta\) in \(0 \leq \theta \leq 2 \pi\) is (a) 2 (b) 5 (c) 6 (d) 12

Find the solution set of the equation, \(\log _{\frac{-x^{2}-6 x}{10}}(\sin 3 x+\sin x)=\log _{\frac{-x^{2}-6 x}{10}}(\sin 2 x)\) Solution: \(\log _{\frac{-x^{2}-6 x}{10}}(\sin 3 x+\sin x)=\log _{\frac{-x^{2}-6 x}{10}}(\sin 2 x)\) \(\Rightarrow \sin 3 x+\sin x>0\) and \(\sin 2 x>0\) \(\Rightarrow 2 \sin 2 x \cos x>0\) and \(\Rightarrow \quad x \in(2 n \pi,(2 n+1) \pi)\) Now, \(\log _{\frac{-x^{2}-6 x}{10}}\left(\frac{\sin 3 x+\sin x}{\sin 2 x}\right)=0\) (By given equation) \(\Rightarrow \frac{2 \sin 2 x \cos x}{\sin 2 x}=1\) \(\Rightarrow 2 \cos x=1\) \(\Rightarrow \quad \cos x=1 / 2 \Rightarrow x=2 n \pi \pm \pi / 3\) \(\Rightarrow x \in(-6,0)\left(\because \frac{-x^{2}-6 x}{10}>0\right.\) and \(\left.\frac{-x^{2}-6 x}{10} \neq 1\right)\) Also \(x \in(2 n \pi,(2 n+1) \pi) \Rightarrow x \in(-2 \pi,-\pi)\) \(\therefore\) for \(n=-1 \Rightarrow x=-2 \pi+\pi / 3=-\frac{5 \pi}{2}\)

Find the general solution of the equtaion, \(2+\tan x\). cot \(\frac{x}{2}+\cot x \cdot \tan \frac{x}{2}=0\) Solution: \(2+\tan x \cdot \cot \frac{x}{2}+\cot x \cdot \tan \frac{x}{2}=0 \quad \ldots(1)\) let \(\tan x . \cot \frac{x}{2}=y\), then equation (1) becomes \(2+\left(y+\frac{1}{y}\right)=0\) \(\Rightarrow y+\frac{1}{y}=-2\) Solving, we get, \(y=-1\) or \(\tan x \cdot \cot \frac{x}{2}=-1\) or \(\frac{\sin x}{\cos x} \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}=-1\)or \(\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{\cos x} \cdot \frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}=-1\) or \(\frac{2 \cos ^{2} \frac{x}{2}}{\cos x}=-1\) or \(\frac{1+\cos x}{\cos x}=-1\) or \(\cos x=-\frac{x}{2}\) or \(x=2 n \pi \pm \frac{2 \pi}{3}\)

If all the solutions ' \(x\) ' of \(a^{\text {oser }}+a^{-\cos x}=6(a>1)\) are real, then set of values of \(a\) is (a) \([3+2 \sqrt{2}, \infty)\) (b) \((6,12)\) (c) \((1,3+2 \sqrt{2})\) (d) None of these
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