Question

Solve the equation arccos(cos(x)) = $\pi -\frac{2x^2}{\pi }$

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation is

$\rm \: arc \: cos(cosx) = \pi \: - \: \dfrac{ {2x}^{2} }{\pi }$

can be rewritten as

$\rm \: {cos}^{ - 1} (cosx) = \pi \: - \: \dfrac{ {2x}^{2} }{\pi }$

We know,

$\rm \: \: {cos}^{ - 1} (cosx) \: = \: x \: \: when \: \: x \: \in \: [0, \: \pi ]$

So,

$\rm \: x \: = \: \pi \: - \: \dfrac{ {2x}^{2} }{\pi }$

$\rm \: x \: = \: \dfrac{ {\pi }^{2} - {2x}^{2} }{\pi }$

$\rm \: {\pi }^{2} - {2x}^{2} = \pi x$

$\rm \: {2x}^{2} + \pi x - {\pi }^{2} = 0$

So, its a quadratic equation, so to find its roots, we use Quadratic Formula.

$\rm \: x = \dfrac{ - ( \pi ) \: \pm \: \sqrt{ {(\pi) }^{2} - 4( - {\pi }^{2})(2)} }{2(2)}$

$\rm \: x = \dfrac{ - \pi \: \pm \: \sqrt{ {\pi }^{2} + 8 {\pi }^{2} } }{4}$

$\rm \: x = \dfrac{ - \pi \: \pm \: \sqrt{ 9 {\pi }^{2} } }{4}$

$\rm \: x = \dfrac{ - \pi \: \pm \: 3\pi }{4}$

$\rm \: x = \dfrac{ - \pi \: + \: 3\pi }{4} \: \: or \: \: \dfrac{ - \pi \: - \: 3\pi }{4}$

$\rm \: x = \dfrac{ 2\pi }{4} \: \: or \: \: \dfrac{ - 4\pi }{4}$

$\rm \: x = \: \dfrac{\pi }{2} \: \: or \: \: - \: \pi \: \: \{rejected \: as \: x \: \in \:[0,\pi ]$

$\rm\implies \:x \: = \:\dfrac{\pi }{2}$

VERIFICATION

Given equation is

$\rm \: arc \: cos(cosx) = \pi \: - \: \dfrac{ {2x}^{2} }{\pi }$

On substituting the value of x, we get

$\rm \: arc \: cos(cos\dfrac{\pi }{2}) = \pi \: - \: \dfrac{ {2\pi }^{2} }{4\pi }$

$\rm \: arc \: cos(0) = \pi \: - \: \dfrac{\pi }{2}$

$\rm \: {cos}^{ - 1}0 \: = \:\dfrac{\pi }{2}$

$\rm\implies \:\dfrac{\pi }{2} \: = \: \dfrac{\pi }{2}$

Hence, Verified

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y = {sin}^{ - 1}(sinx) & \sf  x \: \: if -\dfrac{\pi }{2} \leqslant x \leqslant \dfrac{\pi }{2}\\ \\ \sf y = {cos}^{ - 1}(cosx) & \sf x \: \: if \: 0 \leqslant y \leqslant \pi \\ \\ \sf y = {tan}^{ - 1}(tanx) & \sf x \: \: if \: - \dfrac{\pi }{2} < x < \dfrac{\pi }{2}\\ \\ \sf y = {cosec}^{ - 1}(cosecx) & \sf x \: \: if \: x \: \in \: \bigg[ - \dfrac{\pi}{2}, \: \dfrac{\pi }{2}\bigg] - \{0 \}\\ \\ \sf y = {sec}^{ - 1}(secx) & \sf x \: \: if \: x \: \in \: [0, \: \pi] \: - \: \bigg\{\dfrac{\pi }{2}\bigg\}\\ \\ \sf y = {cot}^{ - 1}(cotx) & \sf x \: \: if \: \:  \in \: \bigg( -  \dfrac{\pi }{2} , \dfrac{\pi }{2}\bigg) - \{0 \} \end{array}} \\ \end{gathered}$

Answer 2

${\pmb{\underline{ \rm{\red{Solution:-}}}}}$

• Answer in the above attachment.

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@Shivam

#BeBrainly