Problem 5
Question
The equation \(\sin ^{4} x-(k+2) \sin ^{2} x-(k+3)=0\) pos- sesses a solution if (A) \(k>-3\) (B) \(k<-2\) (C) \(-3 \leq k \leq-2\) (D) \(k\) is any positive integer
Step-by-Step Solution
Verified Answer
The equation possesses a solution if \(-3 \leq k \leq -2\) (Option C).
1Step 1: Simplify the Equation
We start by simplifying the given equation \( \sin^4 x - (k+2) \sin^2 x - (k+3) = 0 \). Let \( y = \sin^2 x \), transforming the equation into a quadratic: \( y^2 - (k+2)y - (k+3) = 0 \).
2Step 2: Apply the Quadratic Formula
For \( y^2 - (k+2)y - (k+3) = 0 \), we apply the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1 \), \( b = -(k+2) \), \( c = -(k+3) \). This becomes \( y = \frac{k+2 \pm \sqrt{(k+2)^2 + 4(k+3)}}{2} \).
3Step 3: Simplify the Discriminant
Calculate the discriminant: \((k+2)^2 + 4(k+3) = k^2 + 4k + 4 + 4k + 12 = k^2 + 8k + 16\). The square root simplifies to \(\sqrt{k^2 + 8k + 16} = |k+4|\). Thus, \( y = \frac{k+2 \pm |k+4|}{2} \).
4Step 4: Consider Both Cases for Absolute Values
There are two scenarios for \(|k+4|\):1. If \(k+4 \geq 0\), \(y = \frac{k+2 \pm (k+4)}{2}\). Solve for \(y\): - \(y = k+3\) - \(y = -1\) (discard, as \(y \geq 0\)).2. If \(k+4 < 0\), \(y = \frac{k+2 \pm -(k+4)}{2}\). Solve for \(y\): - \(y = -1\) (discard) - \(y = -3\) (discard).
5Step 5: Ensure Solution is Within Valid Range
\( y = \sin^2 x \) implies \(0 \leq y \leq 1\). With \(y = k+3\) from Step 4, set \(0 \leq k+3 \leq 1\). Thus, \(-3 \leq k \leq -2\).
6Step 6: Conclude the Conditions for Solutions
The conditions \(-3 \leq k \leq -2\) mean the equation has solutions only within this range, aligning with option (C).
Key Concepts
Trigonometric EquationsQuadratic TransformationDiscriminant AnalysisInequalities in Trigonometry
Trigonometric Equations
Trigonometric equations involve functions of angles such as sine, cosine, and tangent. These equations often arise in mathematical problems related to wave motion, oscillations, and circular functions.
To solve trigonometric equations, it’s essential to understand the properties and ranges of these functions. For example, the sine function \( \sin x \)has a range between -1 and 1. This means any solution for \( \sin x \) must lie within this interval.
When given an equation like \( \sin^4 x - (k+2) \sin^2 x - (k+3) = 0 \), you'll often start with a substitution to simplify it. Letting \( y = \sin^2 x \) can transform it into a polynomial equation, making it easier to approach with algebraic methods. This substitution helps simplify the complex trigonometric problem into a more familiar form.
To solve trigonometric equations, it’s essential to understand the properties and ranges of these functions. For example, the sine function \( \sin x \)has a range between -1 and 1. This means any solution for \( \sin x \) must lie within this interval.
When given an equation like \( \sin^4 x - (k+2) \sin^2 x - (k+3) = 0 \), you'll often start with a substitution to simplify it. Letting \( y = \sin^2 x \) can transform it into a polynomial equation, making it easier to approach with algebraic methods. This substitution helps simplify the complex trigonometric problem into a more familiar form.
Quadratic Transformation
The process of quadratic transformation involves rewriting a problem—originally in terms of trigonometric functions—into a quadratic equation. This can be a powerful technique to find solutions more easily.
For instance, by letting \( y = \sin^2 x \), the trigonometric equation becomes a standard quadratic equation: \( y^2 - (k+2)y - (k+3) = 0 \). Solving this is straightforward using algebraic methods.
For instance, by letting \( y = \sin^2 x \), the trigonometric equation becomes a standard quadratic equation: \( y^2 - (k+2)y - (k+3) = 0 \). Solving this is straightforward using algebraic methods.
- Use the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- Identify constants: \( a = 1 \), \( b = -(k+2) \), \( c = -(k+3) \)
Discriminant Analysis
Discriminant analysis in quadratic equations plays a crucial role in determining the nature of the roots, whether they are real or complex, and whether they are distinct or repeated.
The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( b^2 - 4ac \). In the context of this problem, analyzing the discriminant \( (k+2)^2 + 4(k+3) \) gives \( k^2 + 8k + 16 \). The square root of this expression simplifies to \( |k+4| \).
Discriminant analysis helps us understand the conditions under which the equation has real solutions:
The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( b^2 - 4ac \). In the context of this problem, analyzing the discriminant \( (k+2)^2 + 4(k+3) \) gives \( k^2 + 8k + 16 \). The square root of this expression simplifies to \( |k+4| \).
Discriminant analysis helps us understand the conditions under which the equation has real solutions:
- If the discriminant is positive, there are two distinct real roots.
- If it is zero, there are two equal real roots.
- If negative, the equation has complex roots.
Inequalities in Trigonometry
Inequalities in trigonometry often require careful interpretation, especially because trigonometric functions have defined ranges. For this problem, where \( y = \sin^2 x \) must satisfy both trigonometric constraints and the given equation, inequalities play a key role.
Since \( \sin^2 x \) is bound between 0 and 1, the solutions derived from the quadratic transformation also need to respect these bounds. In this case, you determine the valid range for \( k \) using the inequalities \( 0 \leq k+3 \leq 1 \).
Therefore, solving the inequality -3 ≤ \( k \) ≤ -2 helps in identifying the values of \( k \) that satisfy both the trigonometric constraint and the solution of the polynomial. This ensures the solutions are realistic given the mathematical and trigonometric contexts. By evaluating these inequalities, you can confidently conclude when the original trigonometric equation possesses solutions.
Since \( \sin^2 x \) is bound between 0 and 1, the solutions derived from the quadratic transformation also need to respect these bounds. In this case, you determine the valid range for \( k \) using the inequalities \( 0 \leq k+3 \leq 1 \).
Therefore, solving the inequality -3 ≤ \( k \) ≤ -2 helps in identifying the values of \( k \) that satisfy both the trigonometric constraint and the solution of the polynomial. This ensures the solutions are realistic given the mathematical and trigonometric contexts. By evaluating these inequalities, you can confidently conclude when the original trigonometric equation possesses solutions.
Other exercises in this chapter
Problem 3
General solution of the equation \((\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2\) is (A) \(2 n \pi \pm \frac{\pi}{4}+\frac{\pi}{12}\) (B) \(n \pi+(-1)^{n
View solution Problem 4
The number of all possible triplets \(\left(a_{1}, a_{2}, a_{3}\right)\) such that \(a_{1}+a_{2} \cos 2 x+a_{3} \sin ^{2} x=0\) for all \(x\) is (A) 0 (B) 1 (C)
View solution Problem 6
The least positive non-integral solution of the equation \(\sin \pi\left(x^{2}+x\right)=\sin \pi x^{2}\) is (A) rational (B) irrational of the form \(\sqrt{p}\)
View solution Problem 7
If \(\sin ^{2} x-2 \sin x-1=0\) has exactly four different solutions in \(x \in[0, n \pi]\), then minimum value of \(n\) can be \((n \in N)\) (A) 4 (B) 3 (C) 2
View solution