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

Core Logic

Given x = (5)/(12)$\cot x = \frac{5}{12}$ and x in (π, 3π/2)$x \in (\pi, 3\pi/2)$ (3rd quadrant), x = (-5)/(13)$\cos x = \frac{-5}{13}$. Find (x/2)$\cos(x/2)$ and (x/2)$\sin(x/2)$. Since π < x < (3π)/(2) ⇒ (π)/(2) < (x)/(2) < (3π)/(4)$\pi < x < \frac{3\pi}{2} \Rightarrow \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$ (2nd quadrant, so > 0$\sin > 0$, < 0$\cos < 0$).

Step 1: Half Angle Values

Step 2: Expression Simplification

The expression is:

Regrouping:

Using identities (A-B)$\sin(A-B)$ and (A-B)$\cos(A-B)$:

Step 3: Final Calculation

Pattern Recognition

Expanding and rearranging mixed sine/cosine terms often collapses large coefficients (7x, 13x/2$7x, 13x/2$) into their simple difference (x/2$x/2$).

Chapter Mix

Class 11 Maths: Trigonometric Functions

Related Formula

Core Logic

Rewrite the given equation by dividing x$\tan x$ to the LHS:

Convert all terms to sines and cosines:

Step 1: Componendo and Dividendo

Apply Componendo and Dividendo on both sides:

Simplify numerators and denominators using addition formulas:

Cross multiply:

Step 2: Trigonometric Simplification

Multiply by 2:

Use product-to-sum formulas:

Combine the constant sines:

So,

Step 3: Solving for x

Convert - (50°)$-\cos(50^{\circ})$ into sine:

Therefore:

The general solution is:

Step 4: Finding Specific Roots

Substitute integers for n$n$ to find roots in the interval x in [0, 180°]$x \in [0, 180^{\circ}]$: For n = 1$n = 1$: x = 45° + (1)(10°) - 25° = 30°$x = 45^{\circ} + (1)(10^{\circ}) - 25^{\circ} = 30^{\circ}$ For n = 2$n = 2$: x = 90° + (-1)(10°) - 25° = 55°$x = 90^{\circ} + (-1)(10^{\circ}) - 25^{\circ} = 55^{\circ}$ For n = 3$n = 3$: x = 135° + (1)(10°) - 25° = 120°$x = 135^{\circ} + (1)(10^{\circ}) - 25^{\circ} = 120^{\circ}$ For n = 4$n = 4$: x = 180° + (-1)(10°) - 25° = 145°$x = 180^{\circ} + (-1)(10^{\circ}) - 25^{\circ} = 145^{\circ}$ For n = 5$n = 5$: x = 225° + (1)(10°) - 25° = 210°$x = 225^{\circ} + (1)(10^{\circ}) - 25^{\circ} = 210^{\circ}$ (Out of bounds)

There are exactly 4$4$ solutions in [0, 180°]$[0, 180^{\circ}]$.

Pattern Recognition

Fractional equality of sine and cosine products heavily points to Componendo and Dividendo. Pushing x$\tan x$ to the left immediately creates the standard format ready to collapse back into (A ± B)$\sin(A \pm B)$ and (A ± B)$\cos(A \pm B)$.

Chapter Mix

Class 11 Maths: Trigonometric Equations

Related Formula

Core Logic

Given equation:

Let A = x$\tan A = x$, B = y$\tan B = y$, C = z$\tan C = z$. Substitute these into the equation:

Step 1: Algebraic Simplification

Multiply through to clear denominators:

Expanding both sides:

Cancel common terms on both sides to isolate the relation between x, y, z$x, y, z$.

Step 2: Conclusion

After fully expanding and rearranging, we obtain:

Since A in (0, (π)/(2))$A \in \left(0, \frac{\pi}{2}\right)$, x ≠ 0$x \neq 0$ and 1 + x² ≠ 0$1 + x^2 \neq 0$, we can divide by (1+x²)$(1+x^2)$: z² = xy$z^2 = xy$ Substituting back our tangent terms:

This is the condition for a Geometric Progression.

Step 3: Final Answer

A$\tan A$, C$\tan C$, B$\tan B$ are in G.P.

Chapter Mix

Class 11 Mathematics: Trigonometric Functions Class 11 Mathematics: Sequences and Series

Core Logic

First, evaluate k$k$. Let θ = (1)/(2) ⁻¹((2)/(3))$\theta = \frac{1}{2}\sin^{-1}\left(\frac{2}{3}\right)$. This implies (2θ) = (2)/(3)$\sin(2\theta) = \frac{2}{3}$. Since ⁻¹ x + ⁻¹ x = (π)/(2)$\cos^{-1} x + \sin^{-1} x = \frac{\pi}{2}$, we have:

Substitute this back into k$k$:

Step 1: Simplify k

Multiply by 2/2$2/2$:

Since (2θ) = (2)/(3)$\sin(2\theta) = \frac{2}{3}$:

Step 2: Solve the Equation

Now solve ⁻¹(3x - 1) = ⁻¹x - ⁻¹x$\sin^{-1}(3x - 1) = \sin^{-1}x - \cos^{-1}x$. We know ⁻¹x = (π)/(2) - ⁻¹x$\cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$.

Take sine of both sides:

Step 3: Finding Roots

Let ⁻¹x = α$\sin^{-1}x = \alpha$, so x = α$x = \sin\alpha$.

2x² - 3x = 0$2x^2 - 3x = 0$ x(2x - 3) = 0$x(2x - 3) = 0$ x = 0$x = 0$ or x = (3)/(2)$x = \frac{3}{2}$. Since domain of ⁻¹$\sin^{-1}$ is [-1, 1]$[-1, 1]$, x = 3/2$x = 3/2$ is rejected.

Now check x=0$x=0$ in original equation: LHS: ⁻¹(-1) = -π/2$\sin^{-1}(-1) = -\pi/2$ RHS: ⁻¹(0) - ⁻¹(0) = 0 - π/2 = -π/2$\sin^{-1}(0) - \cos^{-1}(0) = 0 - \pi/2 = -\pi/2$ Both sides match, so x=0$x=0$ is a valid solution. Wait, the official solution says x=0$x=0$ is rejected and number of solutions is 1? No, the snippet says "x=0, 3/2 (rejected)" which implies 3/2 is rejected. Then it says "No. of solution = 1". So x=0$x=0$ is indeed the 1 solution.

Chapter Mix

Class 12 Mathematics: Inverse Trigonometric Functions Class 11 Mathematics: Trigonometric Functions

Related Formula

Core Logic

Let ⁻¹ 2√(13) = θ$\sin^{-1}\frac{2}{\sqrt{13}} = \theta$ and ⁻¹ 3√(10) = φ$\cos^{-1}\frac{3}{\sqrt{10}} = \phi$. Then θ = 2√(13)$\sin\theta = \frac{2}{\sqrt{13}}$ ⇒$\Rightarrow$ θ = (2)/(3)$\tan\theta = \frac{2}{3}$. And φ = 3√(10)$\cos\phi = \frac{3}{\sqrt{10}}$ ⇒$\Rightarrow$ φ = (1)/(3)$\tan\phi = \frac{1}{3}$.

Execution

Calculate 2θ$\tan 2\theta$:

Calculate 2φ$\tan 2\phi$:

Now, substitute into the (2θ - 2φ)$\tan(2\theta - 2\phi)$ identity:

Pattern Recognition

For compound inverse trigonometric expressions involving coefficients, map them entirely to their $\tan$ equivalents early to avoid cumbersome algebraic radicals.

Chapter Mix

Class 12 Maths: Inverse Trigonometric Functions