If x + ² x = 1$\sin x + \sin^2 x = 1$, x in (0, (π)/(2))$x \in \left(0, \frac{\pi}{2}\right)$, then ( ¹² x + ¹² x) + 3 ( ¹⁰ x + ¹⁰ x + ⁸ x + ⁸ x) + ( ⁶ x + ⁶ x)$(\cos^{12} x + \tan^{12} x) + 3 (\cos^{10} x + \tan^{10} x + \cos^8 x + \tan^8 x) + (\cos^6 x + \tan^6 x)$ is equal to

Related Formula

Core Logic

Given equation: 2 ³ x + 2x x + 4 x - 4 = 0$2 \sin^3 x + \sin 2x \cos x + 4 \sin x - 4 = 0$ Expand 2x$\sin 2x$:

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

Since ² x + ² x = 1$\sin^2 x + \cos^2 x = 1$:

Step 1: Finding appropriate interval for exactly 3 roots

We need exactly 3 solutions in [0, (nπ)/(2)]$\left[0, \frac{n\pi}{2}\right]$. The line y = 2/3$y = 2/3$ intersects the sine wave y = x$y = \sin x$ twice in every 2π$2\pi$ interval. In [0, π]$[0, \pi]$, there are 2 solutions. In [π, 2π]$[\pi, 2\pi]$, there are 0 solutions. In [2π, 3π]$[2\pi, 3\pi]$, there are 2 solutions (total 4 solutions). To get exactly 3 solutions, the interval must stretch past the first root in [2π, 3π]$[2\pi, 3\pi]$, but not reach the second root in that interval. However, the interval is defined as (nπ)/(2)$\frac{n\pi}{2}$. Let's check endpoints (nπ)/(2)$\frac{n\pi}{2}$: For n=4$n=4$: [0, 2π]$[0, 2\pi]$ has 2 solutions. For n=5$n=5$: [0, (5π)/(2)]$[0, \frac{5\pi}{2}]$ includes [2π, 2π + (π)/(2)]$[2\pi, 2\pi + \frac{\pi}{2}]$. Since x = 2/3$\sin x = 2/3$ happens in (0, π/2)$(0, \pi/2)$, there is exactly 1 solution in [2π, 5π/2]$[2\pi, 5\pi/2]$. Thus, total solutions = 3 for n=5$n=5$.

Step 2: Solving quadratic equation

Given n = 5$n = 5$, the quadratic equation is:

Using quadratic formula:

The roots are approximately (-5 ± 4.12)/(2)$\frac{-5 \pm 4.12}{2}$, which evaluates to roughly -0.44$-0.44$ and -4.56$-4.56$. Both roots are strictly negative.

Step 3: Determining interval membership

Since both roots are negative, they belong to the interval (-∞, 0)$(-\infty, 0)$.

Pattern Recognition

Collapsing complex trigonometric expressions often yields c₁ x = c₂$c_1\sin x = c_2$. Overlaying horizontal line intersections on the sine graph bounds n$n$ rapidly by counting nodes.

Chapter Mix

Class 11 Maths: Trigonometric Functions Class 11 Maths: Complex Numbers and Quadratic Equations

Core Logic

Let ex = t$e^{\sin x} = t$, where t > 0$t > 0$ because exponential functions are strictly positive. Substitute into the equation:

Solve for t$t$ using the quadratic formula:

Since t > 0$t > 0$, we discard 1 - √(3)$1 - \sqrt{3}$. Thus, t = 1 + √(3) ≈ 2.732$t = 1 + \sqrt{3} \approx 2.732$. Now, equate back:

We know e ≈ 2.718$e \approx 2.718$. Since 1 + √(3) > e$1 + \sqrt{3} > e$, it follows that ln(1 + √(3)) > 1$\ln(1 + \sqrt{3}) > 1$. But the range of x$\sin x$ is [-1, 1]$[-1, 1]$. Therefore, x$\sin x$ cannot equal a value strictly greater than 1. No real solution exists.

Chapter Mix

Class 11 Maths: Trigonometric Functions Class 12 Maths: Continuity and Differentiability