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

This is a quadratic equation in terms of θ$\csc \theta$. We solve the quadratic roots first and then count the solutions within the given interval.

Step 1: Solve the quadratic equation

Let y = θ$y = \csc \theta$:

Since 16 + 8√(3) = (2 + 2√(3))²$16 + 8\sqrt{3} = (2 + 2\sqrt{3})^2$:

• Case 1 (+$+$ sign):

• Case 2 (-$-$ sign):

Step 2: Count solutions in the interval

Our interval is θ in [-(7π)/(6), (4π)/(3)]$\theta \in \left[-\frac{7\pi}{6}, \frac{4\pi}{3}\right]$:

• For θ = (1)/(2)$\sin \theta = \frac{1}{2}$:

The general solutions are θ = (π)/(6), (5π)/(6)$\theta = \frac{\pi}{6}, \frac{5\pi}{6}$. Within our interval, we have:

• For θ = - √(3)2$\sin \theta = -\frac{\sqrt{3}}{2}$:

The general solutions are θ = -(π)/(3), -(2π)/(3)$\theta = -\frac{\pi}{3}, -\frac{2\pi}{3}$. Within our interval, we have:

Summing the valid solutions:

Pattern Recognition

Perfect Square discriminant: In JEE quadratics with irrational coefficients, the discriminant b²-4ac$b^2-4ac$ almost always simplifies to a perfect square of the form (p + q√(r))²$(p + q\sqrt{r})^2$. Double check your algebraic expansions if it doesn't simplify cleanly.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions

Related Formula

Factorization of quadratic equations by splitting the middle term.

Core Logic

Treat the given equation as a standard quadratic in terms of θ$\cos\theta$ and solve for its roots.

Step 1: Factorization

Split the middle term:

Step 2: Solve for roots

This yields two possible cases:

Step 3: Count Solutions in Interval

The given interval is [-2π, 2π]$[-2\pi, 2\pi]$, which covers two complete cycles of the cosine wave.

• For θ = √(3)2$\cos\theta = \frac{\sqrt{3}}{2}$, there are 2$2$ solutions per cycle 2 × 2 = 4$\implies 2 \times 2 = 4$ solutions.
• For θ = - 1√(2)$\cos\theta = -\frac{1}{\sqrt{2}}$, there are 2$2$ solutions per cycle 2 × 2 = 4$\implies 2 \times 2 = 4$ solutions.

Pattern Recognition

Since both √(3)2$\frac{\sqrt{3}}{2}$ and - 1√(2)$-\frac{1}{\sqrt{2}}$ lie strictly between -1$-1$ and 1$1$, each horizontal line cuts the cosine function exactly twice per period (2π$2\pi$). Across an interval of width 4π$4\pi$, each value must yield exactly 4$4$ solutions.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions

Related Formula

We must reduce the equation to a single trigonometric ratio (like x$\sin x$) using:

Rationalizing fractions:

Core Logic

Simplify the constant term on the RHS:

Now replace ² x = 1 - ² x$\cos^2 x = 1 - \sin^2 x$ in the main equation:

Step 1: Solving the quadratic in x$\sin x$

Let y = x$y = \sin x$:

Thus:

• x = (1)/(2)$\sin x = \frac{1}{2}$
• x = - 2√(3)$\sin x = -\frac{2}{\sqrt{3}}$ (No real solution since |- 2√(3)| ≈ 1.15 > 1$|-\frac{2}{\sqrt{3}}| \approx 1.15 > 1$)

Step 2: Counting solutions in given interval

We must solve x = (1)/(2)$\sin x = \frac{1}{2}$ in x in [-2π, (5π)/(2)] = [-2π, 2.5π]$x \in \left[-2\pi, \frac{5\pi}{2}\right] = [-2\pi, 2.5\pi]$:

• In interval [-2π, 0]$[-2\pi, 0]$: x = -2π + (π)/(6) = -(11π)/(6)$x = -2\pi + \frac{\pi}{6} = -\frac{11\pi}{6}$, x = -2π + (5π)/(6) = -(7π)/(6)$x = -2\pi + \frac{5\pi}{6} = -\frac{7\pi}{6}$ (2 solutions)
• In interval [0, 2π]$[0, 2\pi]$: x = (π)/(6)$x = \frac{\pi}{6}$, x = (5π)/(6)$x = \frac{5\pi}{6}$ (2 solutions)
• In interval [2π, 2.5π]$[2\pi, 2.5\pi]$: x = 2π + (π)/(6) = (13π)/(6)$x = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6}$ (1 solution)

Total number of solutions = 2 + 2 + 1 = 5$2 + 2 + 1 = 5$

Pattern Recognition

Always rationalize standard radical fractions first to find the target integers. Always sketch or trace the sine curve to cross-check solutions across boundaries, particularly at intervals extending slightly beyond multiples of 2π$2\pi$ (like 2.5π$2.5\pi$).

Chapter Mix

Class 11 Trigonometric Functions

Related Formula

Trigonometric compound angle expansion rules:

Core Logic

We need to eliminate the parameter θ$\theta$ between the coordinates of x$x$ and y$y$. Given:

Step 1: Expand y Expression

Now for the y-coordinate:

Step 2: Substitute tan(theta) to Eliminate Parameter

Substitute equation (1) into equation (2):

Matching denominators cancels out, giving:

Alternative expansion matching the standard locus path yields:

Step 3: Match Coefficients and Find Squares Sum

Compare xy - 2√(3)x + 3√(3)y - 6 = 0$xy - 2\sqrt{3}x + 3\sqrt{3}y - 6 = 0$ with the standard form xy + α x + β y + γ = 0$xy + \alpha x + \beta y + \gamma = 0$:

γ = -6$\gamma = -6$

Calculate the sum of squares:

Pattern Recognition

Recognize that (θ + (π)/(3)) - (θ + (π)/(6)) = (π)/(6)$\left(\theta + \frac{\pi}{3}\right) - \left(\theta + \frac{\pi}{6}\right) = \frac{\pi}{6}$, a constant angle. Thus, using (A - B) = ((π)/(6)) = 1√(3)$\tan(A - B) = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ provides a direct shortcut strategy to link x$x$ and y$y$ without fully isolating θ$\tan\theta$.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions Class 11 Mathematics: Straight Lines