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

Intersection method for transcendental configurations: f(x) = g(x)$f(x) = g(x)$ Plot both curves separately to observe distinct intersection markers inside target domain spans.

Core Logic

Rearrange terms to group equations into known standard graphing profiles [cite: 1321]: 3 x = π - 2x x = (π)/(3) - (2x)/(3)$$3\tan x = \pi - 2x \implies \tan x = \frac{\pi}{3} - \frac{2x}{3}$$ [cite: 1321]

Plot the linear equation line y = (π)/(3) - (2x)/(3)$y = \frac{\pi}{3} - \frac{2x}{3}$ along with multiple period tracks of the trigonometric function y = x$y = \tan x$ within the interval [-2π, 2π]$[-2\pi, 2\pi]$[cite: 613, 1321].

Step 1: Point analysis across branches

The line has a negative slope and passes through (0, π/3)$(0, \pi/3)$ and (3π/2, 0)$(3\pi/2, 0)$. Looking across distinct interval chunks separated by asymptotes[cite: 613]:

• Branch 1 (-2π, -(3π)/(2))$\left(-2\pi, -\frac{3\pi}{2}\right)$: 1 intersection
• Branch 2 (-(3π)/(2), -(π)/(2))$\left(-\frac{3\pi}{2}, -\frac{\pi}{2}\right)$: 1 intersection
• Branch 3 (-(π)/(2), (π)/(2))$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$: 1 intersection near origin
• Branch 4 ((π)/(2), (3π)/(2))$\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$: 1 intersection
• Branch 5 ((3π)/(2), 2π)$\left(\frac{3\pi}{2}, 2\pi\right)$: 1 intersection

Counting all distinct points across valid domains gives 5 solutions total[cite: 1337].

Pattern Recognition

A linear curve intersecting tangent asymptote branches will cut exactly once through every continuous range slice unless the line is strictly horizontal or parallel to asymptotes.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions

Related Formula

Trigonometric identity conversion:

Core Logic

Let ²θ = t$\sin^2\theta = t$. Substitute this into the given equation:

Thus, ²θ = (3)/(5)$\sin^2\theta = \frac{3}{5}$ and ²θ = (2)/(5)$\cos^2\theta = \frac{2}{5}$.

Step 1: Simplify Target Expression

Find individual terms from inverse relations: ²θ = (5)/(3) ⁶θ = (125)/(27)$\csc^2\theta = \frac{5}{3} \implies \csc^6\theta = \frac{125}{27}$ ²θ = (5)/(2) ⁶θ = (125)/(8)$\sec^2\theta = \frac{5}{2} \implies \sec^6\theta = \frac{125}{8}$ ⁸θ = ((5)/(2))⁴ = (625)/(16)$\sec^8\theta = \left(\frac{5}{2}\right)^4 = \frac{625}{16}$

Substitute values into expression:

Step 2: Conclusion

Pattern Recognition

Equations structured as A ⁴θ + B ⁴θ = C$A\sin^4\theta + B\cos^4\theta = C$ often yield perfect square trinomial combinations. Check for clean coefficient cancelation steps before computing higher power expressions.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions

Related Formula

Product-to-sum formula and triple angle identity are:

Core Logic

Given equation:

Multiplying by 2:

Using product-to-sum on the first term:

Step 1: Structural Rearrangement

Simplifying through standard trigonometric transformation equations leads directly to:

Hence, either (3θ) = 0$\sin(3\theta) = 0$ or \sin\left(\frac{9\theta}{2}\right) = 0.

Step 2: Finding Roots in the Interval

Interval given:$.

Step 2: Finding Roots in the Interval

Interval given: $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Case A:$.

Case A: $sin(3\theta) = 0 \implies 3\theta = n\pi \implies \theta = \frac{n\pi}{3}Values inside interval:$ Values inside interval: $\left\{-\frac{pi}{3}, 0, \frac{\pi}{3}\right\}(3 solutions).

Case B:$ (3 solutions).

Case B: $sin\left(\frac{9\theta}{2}\right) = 0 \implies \frac{9\theta}{2} = m\pi \implies \theta = \frac{2m\pi}{9}Values inside interval:$ Values inside interval: $\left\{-\frac{4\pi}{9}, -\frac{2\pi}{9}, 0, \frac{2\pi}{9}, \frac{4\pi}{9}\right\}. Since$. Since $0is already counted, this gives 4 unique additional solutions.

Total unique solutions =$ is already counted, this gives 4 unique additional solutions.

Total unique solutions = $3 + 4 = 7.

Pattern Recognition

Transforming powers like$.

Pattern Recognition

Transforming powers like $\cos^3 x$ back into simple multiple-angle terms linearizes trigonometric equations instantly for direct factoring.

Chapter Mix

Class 11 Mathematics: Trigonometry

Related Formula

Logarithmic addition property: (a) + (b) = (ab)$\log(a) + \log(b) = \log(ab)$. Double angle sine formula: 2 x x = 2x$2\sin x \cos x = \sin 2x$.

Step 1: Simplify Set A

Combine the logarithmic elements:

Since π² ≈ 9.87$\pi^2 \approx 9.87$, (8)/(π²) ≈ 0.81$\frac{8}{\pi^2} \approx 0.81$, which is less than 1. Plotting | 2x| = (8)/(π²)$|\sin 2x| = \frac{8}{\pi^2}$ over the specified range x in (0, π)$x \in (0, \pi)$ yields exactly 4 real intersection points.

Hence, n(A) = 4$n(A) = 4$.

Step 2: Simplify Set B

Let √(x) = t$\sqrt{x} = t$ where t ≥ 0$t \ge 0$. The equation becomes:

Case I: If t < 2$t < 2$ :

Case II: If t > 2$t > 2$ :

Hence, set B = 0, 1, 9, 16$B = \{0, 1, 9, 16\}$, giving n(B) = 4$n(B) = 4$.

Step 3: Calculate Union

Since all elements of set A are non-integral angles in (0, π)$(0, \pi)$ and elements of set B are pure integers, the sets are completely disjoint (A B =$A \cap B = \emptyset$).

Pattern Recognition

Always separate functions into disjoint numeric domains (e.g., angles vs. whole integers) to conclude unions without performing tedious tracking of individual values manually.

Chapter Mix

Class 11 Physics: Trigonometric Functions Class 11 Mathematics: Sets

Related Formula

Identity for reciprocal of product of sines with an arithmetic progression phase difference β$\beta$:

where β = A - B$\beta = A - B$.

Core Logic

Let θᵣ = (π)/(4) + r(π)/(6)$\theta_r = \frac{\pi}{4} + r\frac{\pi}{6}$. Then the difference between consecutive angles is:

Multiply and divide the general term of the summation by ((π)/(6))$\sin\left(\frac{\pi}{6}\right)$:

Step 1: Expand the Telescopic Sum

Where:

Now compute ((23π)/(12)) = (2π - (π)/(12)) = - ((π)/(12))$\cot\left(\frac{23\pi}{12}\right) = \cot\left(2\pi - \frac{\pi}{12}\right) = -\cot\left(\frac{\pi}{12}\right)$:

Step 2: Solve for a and b

Accounting for the absolute value ranges across the quadrants and simplifying the telescopic intervals:

Comparing with a√(3) + b$a\sqrt{3} + b$:

Now calculate a² + b²$a^2 + b^2$:

Pattern Recognition

Telescopic series involving absolute values of trigonometric products require careful tracking of interval signs across quadrants before applying boundary difference reductions.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions