Question

6-4sin4x-6cos8x=9sec^2theta+cosec^2 theta;t x1 be smallest +ve sol and x2 be

Answer 1

x_1=\dfrac{3\pi}{8} and x_2=\dfrac{7\pi}{8} with \theta=\dfrac{\pi}{6}

Answer 2

We start with the equation

$$6-4\sin4x-6\cos8x=9\sec^2\theta+\csc^2 \theta.$$

Step 1. Simplify the Left‐Hand Side (in $x$)

Write $\cos8x$ in terms of $\sin4x$ by using the double‐angle formula:

$$\cos8x=1-2\sin^2 4x.$$

Then

$$\begin{aligned} 6-4\sin4x-6\cos8x &=6-4\sin4x-6(1-2\sin^2 4x)\\[1mm] &=6-4\sin4x-6+12\sin^2 4x\\[1mm] &=12\sin^2 4x-4\sin4x\\[1mm] &=4\sin4x\,(3\sin4x-1). \end{aligned}$$

Step 2. Understand the Right‐Hand Side (in $\theta$)

We need to note that

$$\sec^2\theta=\frac{1}{\cos^2\theta},\quad \csc^2\theta=\frac{1}{\sin^2\theta}.$$

Thus

$$9\sec^2\theta+\csc^2\theta=\frac{9}{\cos^2\theta}+\frac{1}{\sin^2\theta}.$$

One may show (by the substitution $t=\tan\theta$) that this expression is minimized with respect to $\theta$ when

$$\tan^2\theta=\frac{1}{3}.$$

Indeed, writing

$$\frac{9}{\cos^2\theta}+\frac{1}{\sin^2\theta}=(1+t^2)\Bigl(9+\frac{1}{t^2}\Bigr)=9t^2+\frac{1}{t^2}+10,$$

minimization gives $t^2=\frac{1}{3}$ and the minimum value is

$$9\left(\frac{1}{3}\right)+3+10=3+3+10=16.$$

Thus

$$9\sec^2\theta+\csc^2\theta\ge 16,$$

with equality when $\tan^2\theta=\frac{1}{3}$ (i.e. $\theta=\frac{\pi}{6}$ in the first quadrant).

Step 3. Equate the Two Sides at the Minimum

For the equality to hold the left‐hand side must also equal 16:

$$4\sin4x\,(3\sin4x-1)=16.$$

Dividing both sides by 4,

$$\sin4x\,(3\sin4x-1)=4.$$

Let $s=\sin4x$. Then

$$3s^2 - s -4=0.$$

The quadratic has discriminant

$$\Delta=1+48=49,$$

so

$$s=\frac{1\pm 7}{6}.$$

This gives:

$$s=\frac{8}{6}=\frac{4}{3}\quad \text{or}\quad s=\frac{-6}{6}=-1.$$

Since $\sin4x$ must lie in $[-1,1]$, we reject $s=\frac{4}{3}$. Thus we have

$$\sin4x=-1.$$

Step 4. Solve $\sin4x=-1$

We know

$$\sin\alpha=-1 \quad\Longrightarrow\quad \alpha=\frac{3\pi}{2}+2\pi k,\quad k\in\mathbb{Z}.$$

Since $\alpha=4x$, it follows that

$$4x=\frac{3\pi}{2}+2\pi k,$$

or

$$x=\frac{3\pi}{8}+\frac{\pi k}{2},\quad k\in\mathbb{Z}.$$

Thus the smallest positive solution is (with $k=0$)

$$x_1=\frac{3\pi}{8},$$

and the next positive solution (with $k=1$) is

$$x_2=\frac{3\pi}{8}+\frac{\pi}{2}=\frac{3\pi+4\pi}{8}=\frac{7\pi}{8}.$$

Also, from before, equality on the right‐hand side is reached when

$$\tan\theta=\frac{1}{\sqrt{3}},\quad\text{so}\quad \theta=\frac{\pi}{6}.$$

Minimal Core Explanation

1. Simplify LHS:
   $\cos8x=1-2\sin^2 4x$ gives $$6-4\sin4x-6(1-2\sin^2 4x)=4\sin4x(3\sin4x-1).$$
2. Minimize RHS:
   Show that $$9\sec^2\theta+\csc^2\theta\ge16,$$ with equality when $\tan^2\theta=\frac{1}{3}$ (i.e. $\theta=\frac{\pi}{6}$).
3. Setting equality:
   $4\sin4x(3\sin4x-1)=16$ leads to $$\sin4x(3\sin4x-1)=4.$$ Let $s=\sin4x$. Solving $3s^2-s-4=0$ gives $s=-1$ (since $s=\frac{4}{3}$ is not possible).
4. Solve for $x$:
   $\sin4x=-1$ yields $$x=\frac{3\pi}{8}+\frac{\pi k}{2}.$$ Thus $x_1=\frac{3\pi}{8}$ (smallest positive) and $x_2=\frac{7\pi}{8}$ (next solution).