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a) $p^2-8>0$

b) $p=\sqrt 2$

c) $p=-\sqrt 2$

d) $p=0$

My attempt

$$\frac{\sec x +\csc x}{2} \ge \sqrt {\sec x \csc x}$$

$$\frac{\sin x +\cos x}{\sin x \cos x }\ge 2\sqrt {\frac{1}{\sin x \cos x}}$$

$$\sin x +\cos x \ge 2\sqrt {\sin x \cos x}$$

$$(\sqrt {\sin x}-\sqrt {\cos x})^2\ge 0$$

$$\sin x \ge \cos x$$

I realise that some of that squaring might have removed or added some roots, but I don’t know what else to do

From this result, the interval for $x$ is $[\frac{\pi}{4}, \pi]\cup [\frac{5\pi}{4}, \frac{3\pi}{2}]$

Don’t know what do next. Can I get some insight?

Another attempt

$$\sin x +\cos x =p \sin x \cos x$$ $$1+2\sin x\cos x =\frac 14 p^2 4\sin^2x \cos ^2x$$

$$1+\sin 2x =\frac 14 p\sin^2 2x$$

$$p\sin^22x-4\sin 2x -4=0$$

Community
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Aditya
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3 Answers3

1

Squaring an equation may introduce unwanted roots, so it is generally avoided. Rewrite the equation, instead, as $$p(x) = \frac{\sin x +\cos x}{\frac12\sin 2x} =\frac{2\sqrt2 \cos(x-\frac\pi4)}{2\cos^2(x-\frac \pi4)-1}$$

So, $p(x)$ is a function of $\cos(x-\frac\pi4)$ and symmetric around $\frac\pi4$ and $\frac{5\pi}4$, which are also the two local extrema over $(0,2\pi)$, i.e. $$p_{min}(\frac\pi4) = 2\sqrt2,\>\>\>\>\>p_{max}(\frac{5\pi}4) = -2\sqrt2$$

Thus, for any $p$ satisfying $p^2>(\pm 2\sqrt2)^2=8$, it meets the curve $p=\sec x+ \csc x$ four times, i.e. four distinctive roots in the shaded area of the graph below,

enter image description here

Quanto
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  • How did you find the extremes, because graphs won’t be accessible. I think derivatives will only give a maxima or a minima – Aditya Apr 21 '20 at 04:15
  • @Aditya - note that $p(x)$ has vertical asymptotes at $2\cos^2(x-\frac\pi4)$, or $0, >\frac\pi2, >\pi,>\frac{3\pi}2,> 2\pi$. For these 4 domains, it can be analyzed that $p(x) \ge 2\sqrt2,> x\in(0,\frac\pi2)$; $p(x) \le -2\sqrt2,> x\in(\pi,\frac{3\pi}2)$; $p(x) \in R,> x\in(\frac\pi2,\pi) \cup (\frac{3\pi}2,2\pi)$ – Quanto Apr 21 '20 at 04:36
  • I tried a different way to solve it. Can you please check the edit? – Aditya May 31 '20 at 07:20
  • @Aditya - I would not square the equation, which introduces additional solutions that need to be excluded at the end – Quanto May 31 '20 at 19:20
  • @Quanto-Isn't there a flaw in which the options are given. If $p^2=9$ then equation has actually 8 solutions. Four for $p=-3$ and another four for $p=3$ – Maverick Feb 27 '21 at 03:06
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    @Maverick - It may be a bit subtle the way the question is asked, “which is incorrect?”. You could then justify a) – Quanto Feb 27 '21 at 17:44
0

$$\sin x+\cos x=p\cos x\sin x$$

Now if $y=\sin x+\cos x=\sqrt2\sin(\pi/4+x),y^2\le2$

$$2y=p(y^2-1)\iff py^2-2y-p=0$$

As the discriminant $>0,$ there will always be four real roots in$(0,2\pi)$

Observe that if $z=t$ is a solution of $$\sin z=a$$ so will be $z=\pi-t$

Now $t,\pi-t$ will coincide if $t=2m\pi+\pi-t\implies\cos2t=-1$

So, here $$1-y^2=-1\iff y=?$$

To coincide, we need $$p(2-1)=2\cdot\pm\sqrt2$$

lab bhattacharjee
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0

The derivative of $p$ is equal to $\dfrac{\sin(x)}{\cos^2(x^2)}-\dfrac{\cos(x)}{\sin^2(x)}$ so$p$ has a minimum at $x=\dfrac{\pi}{4}$ and a maximum at $x=\dfrac{5\pi}{4}$. Consequently $p$ has four values in $[0,2\pi]$ for $\sec x + \csc x \gt2\sqrt2$ and for $\sec x + \csc x \lt-2\sqrt2$.

Thus only $p^2-8\gt0$ is correct.

Piquito
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