If 2 sin³ x + sin 2x cos x + 4 sin x - 4 = 0 has exactly 3 solutions in the interval [0, nπ/2], n ∈

Q: Wait, why can't we just divide by $\sin x$ at the beginning to simplify?

That is a classic trap! Dividing by a variable expression like $\sin x$ assumes that $\sin x \neq 0$. If you do that, you might lose potential solutions where $\sin x = 0$. Always factor instead of dividing to ensure you capture all possible roots.

Q: How do I know which solutions to include in the interval $[0, \frac{n\pi}{2}]$?

The best way is to sketch the graph of $y = \sin x$ and the line $y = \frac{2}{3}$. The solutions appear in pairs within each $2\pi$ cycle: $\alpha$ and $\pi - \alpha$ in the first cycle, and $2\pi + \alpha$ and $3\pi - \alpha$ in the second. By placing the interval $[0, \frac{n\pi}{2}]$ on the x-axis, you can visually see which solutions fall inside.

Q: Why is $n$ an integer? The problem says $n \in \mathbb{N}$.

The problem specifies $n \in \mathbb{N}$, which means $n$ must be a natural number (a positive integer). Our inequality $4.46 \le n < 5.54$ restricts $n$ to a range where only one integer, $n = 5$, satisfies the condition of having exactly 3 solutions.

Question

If $2 \sin^3 x + \sin 2x \cos x + 4 \sin x - 4 = 0$ has exactly 3 solutions in the interval $[0, \frac{n\pi}{2}]$, $n \in \mathbb{N}$, then the roots of the equation $x^2 + nx + (n - 3) = 0$ belong to :

Options:

(A) $(0, \infty)$

(B) $(-\infty, 0)$

(C) $(-\frac{\sqrt{17}}{2}, \frac{\sqrt{17}}{2})$

(D) $\mathbb{Z}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given trigonometric equation is:

$$2 \sin^3 x + \sin 2x \cos x + 4 \sin x - 4 = 0$$

To simplify this, we apply the double angle identity, $\sin 2x = 2 \sin x \cos x$. Substituting this into the second term yields $(2 \sin x \cos x) \cos x$, which simplifies to $2 \sin x \cos^2 x$.

The equation now becomes:

$$2 \sin^3 x + 2 \sin x \cos^2 x + 4 \sin x - 4 = 0$$

### The Master Equation

We factor out $2 \sin x$ from the first two terms:

$$2 \sin x (\sin^2 x + \cos^2 x) + 4 \sin x - 4 = 0$$

Using the Pythagorean identity, $\sin^2 x + \cos^2 x = 1$, the expression collapses significantly. The equation reduces to:

$$2 \sin x (1) + 4 \sin x - 4 = 0$$

$$6 \sin x = 4 \Rightarrow \sin x = \frac{2}{3}$$

### Visualizing the Solutions

With $\sin x = \frac{2}{3}$, we identify the solutions by considering the intersection of $y = \sin x$ and $y = \frac{2}{3}$. Let $\alpha = \sin^{-1}(\frac{2}{3})$.

The solutions are located at:

$$x_1 = \alpha, \quad x_2 = \pi - \alpha, \quad x_3 = 2\pi + \alpha, \quad x_4 = 3\pi - \alpha$$

### The Interval Trap

The problem requires exactly 3 solutions in the interval $[0, \frac{n\pi}{2}]$. To satisfy this, the interval must contain $x_1, x_2,$ and $x_3$, but must exclude $x_4$.

This imposes the condition:

$$x_3 \le \frac{n\pi}{2} < x_4$$

Substituting the values of $x_3$ and $x_4$:

$$2\pi + \alpha \le \frac{n\pi}{2} < 3\pi - \alpha$$

Dividing by $\frac{\pi}{2}$, we obtain:

$$4 + \frac{2\alpha}{\pi} \le n < 6 - \frac{2\alpha}{\pi}$$

Given $\sin \alpha = \frac{2}{3}$, $\alpha \approx 0.73$ radians, meaning $\frac{2\alpha}{\pi} \approx 0.46$. Thus, $4.46 \le n < 5.54$. The only natural number satisfying this is **$n = 5$**.

### The Final Quadratic Analysis

With $n = 5$, the quadratic equation is $x^2 + 5x + (5 - 3) = 0$, which simplifies to:

$$x^2 + 5x + 2 = 0$$

Applying the quadratic formula:

$$x = \frac{-5 \pm \sqrt{25 - 8}}{2} = \frac{-5 \pm \sqrt{17}}{2}$$

Since $\sqrt{17} \approx 4.12$, both roots are negative:

$$x_1 = \frac{-5 + 4.12}{2} \approx -0.44, \quad x_2 = \frac{-5 - 4.12}{2} \approx -4.56$$

Both roots lie in the interval $(-\infty, 0)$. You have successfully navigated the trigonometry, constrained the interval, and solved the quadratic.

Animated Solution for Mathematics - Trigonometry: If $2 sin^{3} x + sin 2 x cos x + 4 sin x - 4 = 0$ has exactly 3 solutions in the interval $[0, \frac{nπ}{2}]$ , $n \in N$ , then the roots of the equation $x^{2} + n x + (n - 3) = 0$ belong to :

Select Answer:

Visualized Solution

The Trigonometric Equation

Double Angle Identity

Substitution and Factoring

Applying Pythagorean Identity

Solving for $sin x$

Visualizing the Solutions

Identifying the First Two Solutions

Identifying the Next Solutions

Setting the Interval Condition

Solving the Inequality for $n$

Finding the Integer $n$

The Quadratic Equation

Finding the Roots

Analyzing the Roots

The Sigma Insight: General Solution of Trigonometric Equations

Analyzing the Setup

The Master Equation

Visualizing the Solutions

The Interval Trap

The Final Quadratic Analysis

Similar Questions

The number of solutions of $sin^{2} x + (2 + 2 x - x^{2}) sin x - 3 (x - 1)^{2} = 0$ , where $- π \leq x \leq π$ , is________

The number of solutions of the equation $4 sin^{2} x - 4 cos^{3} x + 9 - 4 cos x = 0$ ; $x \in [- 2 π, 2 π]$ is :

The number of solutions of equation $(4 - 3) sin x - 23 cos^{2} x = - \frac{4}{1 + 3}, x \in [- 2 π, \frac{5 π}{2}]$ is

Find all the solution of $4 cos^{2} x sin x - 2 sin^{2} x = 3 sin x$

The number of solutions of equation $(4 - 3) sin x - 23 cos^{2} x = - \frac{4}{1 + 3}, x \in [- 2 π, \frac{5 π}{2}]$ is

Find all the solution of $4 cos^{2} x sin x - 2 sin^{2} x = 3 sin x$ .

For $x \in (0, π)$ , the equation $sin x + 2 sin 2 x - sin 3 x = 3$ has

The number of values of $x$ in the interval $[0, 3 π]$ satisfying the equation $2 sin^{2} x + 5 sin x - 3 = 0$ is

The number of solutions of the equation $1 + sin^{4} x = cos^{2} 3 x$ , $x \in [- \frac{5 π}{2}, \frac{5 π}{2}]$ is :

The number of solutions of the equation $2 x + 3 tan x = π, x \in [- 2 π, 2 π] - {\pm \frac{π}{2}, \pm \frac{3 π}{2}}$ is

Select Answer:

Visualized Solution

The Trigonometric Equation

Double Angle Identity

Substitution and Factoring

Applying Pythagorean Identity

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinx

Visualizing the Solutions

Identifying the First Two Solutions

Identifying the Next Solutions

Setting the Interval Condition

Solving the Inequality for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Finding the Integer .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

The Quadratic Equation

Finding the Roots

Analyzing the Roots

The Sigma Insight: General Solution of Trigonometric Equations

Analyzing the Setup

The Master Equation

Visualizing the Solutions

The Interval Trap

The Final Quadratic Analysis

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of solutions of sin2x+(2+2x−x2)sinx−3(x−1)2=0, where −π≤x≤π, is________

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of solutions of the equation 4sin2x−4cos3x+9−4cosx=0; x∈[−2π,2π] is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of solutions of equation (4−3​)sinx−23​cos2x=−1+3​4​,x∈[−2π,25π​] is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find all the solution of 4cos2xsinx−2sin2x=3sinx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of solutions of equation (4−3​)sinx−23​cos2x=−1+3​4​,x∈[−2π,25π​] is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find all the solution of 4cos2xsinx−2sin2x=3sinx.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For x∈(0,π), the equation sinx+2sin2x−sin3x=3 has

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of values of x in the interval [0,3π] satisfying the equation 2sin2x+5sinx−3=0 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of solutions of the equation 1+sin4x=cos23x, x∈[−25π​,25π​] is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of solutions of the equation 2x+3tanx=π,x∈[−2π,2π]−{±2π​,±23π​} is

Solving for $sin x$

Solving the Inequality for $n$

Finding the Integer $n$

The number of solutions of $sin^{2} x + (2 + 2 x - x^{2}) sin x - 3 (x - 1)^{2} = 0$ , where $- π \leq x \leq π$ , is________

The number of solutions of the equation $4 sin^{2} x - 4 cos^{3} x + 9 - 4 cos x = 0$ ; $x \in [- 2 π, 2 π]$ is :

The number of solutions of equation $(4 - 3) sin x - 23 cos^{2} x = - \frac{4}{1 + 3}, x \in [- 2 π, \frac{5 π}{2}]$ is

Find all the solution of $4 cos^{2} x sin x - 2 sin^{2} x = 3 sin x$

The number of solutions of equation $(4 - 3) sin x - 23 cos^{2} x = - \frac{4}{1 + 3}, x \in [- 2 π, \frac{5 π}{2}]$ is

Find all the solution of $4 cos^{2} x sin x - 2 sin^{2} x = 3 sin x$ .

For $x \in (0, π)$ , the equation $sin x + 2 sin 2 x - sin 3 x = 3$ has

The number of values of $x$ in the interval $[0, 3 π]$ satisfying the equation $2 sin^{2} x + 5 sin x - 3 = 0$ is

The number of solutions of the equation $1 + sin^{4} x = cos^{2} 3 x$ , $x \in [- \frac{5 π}{2}, \frac{5 π}{2}]$ is :

The number of solutions of the equation $2 x + 3 tan x = π, x \in [- 2 π, 2 π] - {\pm \frac{π}{2}, \pm \frac{3 π}{2}}$ is