Let A=xin(0,π)-(π)/(2): (2/π)| x|+ (2/π)| x|=2$A=\left\{x\in(0,\pi)-\left\{\frac{\pi}{2}\right\}:\log_{(2/\pi)}|\sin x|+\log_{(2/\pi)}|\cos x|=2\right\}$ and B=x≥0:√(x)(√(x)-4)-3|√(x)-2|+6=0.$B=\left\{x\ge0:\sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\right\}.$ Then n(A B)$n(A\cup B)$ is equal to:

Related Formula

Trigonometric Sine Identity:

Core Logic

Let ⁻¹x = θ x = θ$\sin^{-1}x = \theta \implies x = \sin\theta$ and √(1-x²) = θ$\sqrt{1-x^2} = \cos\theta$. Given constraint -(1)/(2) < x < 1√(2) -(π)/(6) < θ < (π)/(4)$-\frac{1}{2} < x < \frac{1}{\sqrt{2}} \implies -\frac{\pi}{6} < \theta < \frac{\pi}{4}$.

Substitute parameter representations into expression:

Step 1: Check Principal Bounds

Evaluate bounds for arguments: since -(π)/(6) < θ < (π)/(4)$-\frac{\pi}{6} < \theta < \frac{\pi}{4}$:

This lies completely within the principal value branch of ⁻¹x$\sin^{-1}x$, which is [-(π)/(2), (π)/(2)]$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, ⁻¹( (θ + (π)/(6))) = θ + (π)/(6)$\sin^{-1}\left(\sin\left(\theta + \frac{\pi}{6}\right)\right) = \theta + \frac{\pi}{6}$.

Step 2: Final Form

Substituting back θ = ⁻¹x$\theta = \sin^{-1}x$:

Pattern Recognition

Always check primary interval bounds when stripping inverse operators. If the arguments exceed bounds, quadrant mapping transformations must be performed.

Chapter Mix

Class 12 Mathematics: Inverse 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

The standard conversion between ⁻¹(x)$\cot^{-1}(x)$ and ⁻¹(x)$\tan^{-1}(x)$ depends on the sign of x$x$:

Core Logic

Simplify the interior terms algebraic representations:

Step 1: Convert to Inverse Tangent terms

Since α > β > γ > 0$\alpha > \beta > \gamma > 0$:

• (1+αβ)/(α-β) > 0 ⇒ ⁻¹((1+αβ)/(α-β)) = ⁻¹((α-β)/(1+αβ))$\frac{1+\alpha\beta}{\alpha-\beta} > 0 \Rightarrow \cot^{-1}\left(\frac{1+\alpha\beta}{\alpha-\beta}\right) = \tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right)$
• (1+βγ)/(β-γ) > 0 ⇒ ⁻¹((1+βγ)/(β-γ)) = ⁻¹((β-γ)/(1+βγ))$\frac{1+\beta\gamma}{\beta-\gamma} > 0 \Rightarrow \cot^{-1}\left(\frac{1+\beta\gamma}{\beta-\gamma}\right) = \tan^{-1}\left(\frac{\beta-\gamma}{1+\beta\gamma}\right)$
• (1+αγ)/(γ-α) < 0$\frac{1+\alpha\gamma}{\gamma-\alpha} < 0$ (since γ - α < 0$\gamma - \alpha < 0$) ⇒ ⁻¹((1+αγ)/(γ-α)) = π + ⁻¹((γ-α)/(1+αγ))$\Rightarrow \cot^{-1}\left(\frac{1+\alpha\gamma}{\gamma-\alpha}\right) = \pi + \tan^{-1}\left(\frac{\gamma-\alpha}{1+\alpha\gamma}\right)$

Step 2: Telescopic Sum Evaluation

Apply the difference identity for arctan, ⁻¹((x-y)/(1+xy)) = ⁻¹x - ⁻¹y$\tan^{-1}\left(\frac{x-y}{1+xy}\right) = \tan^{-1}x - \tan^{-1}y$ :

All variables cancel symmetrically leaving:

= π$= \pi$

Pattern Recognition

The sign trap is the most vital component of this question. The ordering α > β > γ > 0$\alpha > \beta > \gamma > 0$ means the last term contains a denominator with a negative difference (γ - α)$(\gamma - \alpha)$, introducing the +π$+\pi$ offset according to the principal range of ⁻¹(x)$\cot^{-1}(x)$.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions