SATHEE: Trigonometrical Equations 3 Question 16

Trigonometrical Equations 3 Question 16

17. Find the smallest positive number $p$ for which the equation $\cos (p \sin x) = \sin (p \cos x)$ has a solution $x \in [0, 2 π]$ .

(1995, 5M)

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Answer:

Correct Answer: 17. Smallest positive value of $p = \frac{π}{2 \sqrt{2}}$

Solution:

Given, $\cos (p \sin x) = \sin (p \cos x), \forall x \in [0, 2 π]$

$\begin{aligned} \Rightarrow & \cos (p \sin x) & = \cos \frac{π}{2} - p \cos x \Rightarrow & p \sin x & = 2 n π \pm \frac{π}{2} - p \cos x, n \in I \end{aligned}$

$\Rightarrow p \sin x + p \cos x = 2 n π + π / 2$

or $p \sin x - p \cos x = 2 n π - π / 2, n \in I$

$\Rightarrow p (\sin x + \cos x) = 2 n π + π / 2$

or $p (\sin x - \cos x) = 2 n π - π / 2, n \in I$

$\Rightarrow p \sqrt{2} (\cos \frac{π}{4} \sin x + \sin \frac{π}{4} \cos x) = 2 n π + \frac{π}{2}$

or $p \sqrt{2} \cos \frac{π}{4} \sin x - \sin \frac{π}{4} \cos x = 2 n π - \frac{π}{2}, n \in I$

$\Rightarrow p \sqrt{2} [\sin (x + π / 4)] = \frac{(4 n + 1) π}{2}$

$p \sqrt{2} [\sin (x - π / 4)] = (4 n - 1) \frac{π}{2}, n \in I$

Now, $- 1 \leq \sin (x \pm π / 4) \leq 1$

$\Rightarrow - p \sqrt{2} \leq p \sqrt{2} \sin (x \pm π / 4) \leq p \sqrt{2}$

$\Rightarrow - p \sqrt{2} \leq \frac{(4 n + 1) \cdot π}{2} \leq p \sqrt{2}, n \in I$

or $- p \sqrt{2} \leq \frac{(4 n - 1) π}{2} \leq p \sqrt{2}, n \in I$

Second inequality is always a subset of first, therefore we have to consider only first.

It is sufficient to consider $n \geq 0$ , because for $n > 0$ , the solution will be same for $n \geq 0$ .

If $n \geq 0$

$- \sqrt{2} p \leq (4 n + 1) π / 2$

$\Rightarrow (4 n + 1) π / 2 \leq \sqrt{2} p$

For $p$ to be least, $n$ should be least.

$\begin{array}{ll} \Rightarrow & n = 0 \\ \Rightarrow & \sqrt{2} p \geq π / 2 \Rightarrow p \geq \frac{π}{2 \sqrt{2}} \end{array}$

Therefore, least value of $p = \frac{π}{2 \sqrt{2}}$

Trigonometrical Equations 3 Question 16

17. Find the smallest positive number $p$ for which the equation $\cos (p \sin x) = \sin (p \cos x)$ has a solution $x \in [0, 2 π]$ .

Answer:

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Trigonometrical Equations 3 Question 16

17. Find the smallest positive number p for which the equation cos⁡(psin⁡x)=sin⁡(pcos⁡x) has a solution x∈[0,2π].

Answer:

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17. Find the smallest positive number $p$ for which the equation $\cos (p \sin x) = \sin (p \cos x)$ has a solution $x \in [0, 2 π]$ .