If $ x \in \left(0,\frac{\pi}{2}\right)$. Then value of $x$ in $ \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$

Question

If $\displaystyle x \in \left(0,\frac{\pi}{2}\right)$ then find a value of $x$ in $\displaystyle \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$

$\bf{Attempt:}$ From $$\frac{3}{\sqrt{2}\cos x}-\frac{\sqrt{2}}{\sin x} = 1.$$

$$3\sin x-2\cos x = \sqrt{2}\sin x\cos x$$

$$(3\sin x-2\cos x)^2 = 2\sin^2 x\cos^2 x$$

$$9\sin^2 x+4\cos^2 x-12 \sin x\cos x = 2\sin^2 x\cos^2 x$$

Could some help me to solve it, thanks

Answer 1

Use $\sin{x}=\frac{2t}{1+t^2}$ and $\cos{x}=\frac{1-t^2}{1+t^2}$, where $t=\tan\frac{x}{2}$.

One of roots is $\sqrt2-1$ and the second is very ugly.

For $\tan\frac{x}{2}=\sqrt2-1$ we obtain $\frac{x}{2}=22.5^{\circ}+180^{\circ}k,$ where $k\in\mathbb Z$,

which gives $x=\frac{\pi}{4}$.

The second root does not give solutions.

Answer 2

The function is increasing in $\left(0, \dfrac{\pi}{2}\right)$ which can be seen easily by differentiating and seeing that the derivative is positive (in the given interval).

It is easy to see that $x = \dfrac{\pi}{4}$ is a root and so is the only root in $\left(0, \dfrac{\pi}{2}\right)$

Answer 3

I think it's better to make the following.

Let $x=\frac{\pi}{4}+t$, where $t\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$.

Hence, we need to solve that $$3\sin{x}-2\cos{x}=\sqrt2\sin{x}\cos{x}$$ or $$3(\sin{t}+\cos{t})-2(\cos{t}-\sin{t})=\cos^2t-\sin^2t$$ or $$\sin{t}(5+\sin{t})+(1-\cos{t})\cos{t}=0$$ or $$\sin\frac{t}{2}\left(\cos\frac{t}{2}(5+\sin{t})+\sin\frac{t}{2}\cos{t}\right)=0$$ and since $$\cos\frac{t}{2}(5+\sin{t})+\sin\frac{t}{2}\cos{t}=5\cos\frac{t}{2}+\sin\frac{3x}{2}>5\cos\frac{\pi}{8}-1>0,$$ we obtain $\sin\frac{t}{2}=0$, which gives $t=0$ and $x=\frac{\pi}{4}$.

Answer 4

To take advantage of $\sqrt2$ in simplification restructure the equation to:

$$\frac{3 \cdot \dfrac{1}{\sqrt{2}}}{\cos x}-\frac{2 \cdot \dfrac{1}{\sqrt{2}}}{\sin x} = 1,$$

and so by inspection it is obviously $ x= \pi/4.$