Question

find the coordinates of the point of intersection of the curves y = cos x, y= sin3x
$if \: - \frac{π}{2} \leqslant x \leqslant \frac{π}{2}$
​

Answer 1

$\large\textsf{The point of intersection is given by}$

$\sf{ \sin3x = \cos \: x = \sin} \bigg\lgroup \sf{ \frac{π}{2} - x}\bigg\rgroup$

$\implies \sf{3x = n\pi + ( - {1)}^{n}} \bigg\lgroup \sf{ \frac{π}{2} - x}\bigg\rgroup$

$\sf{ \clubs \: \: Let \: \: n \: \: be \: \: even \: \: i.e. \: n = 2m}$

$\implies \sf{3x = 2m\pi + } \bigg\lgroup \sf{ \frac{π}{2} - x}\bigg\rgroup$

$\implies \sf {\pink{x = \frac{mn}{2} + \frac{\pi}{8} }} \: ....(i)$

$\sf{} \clubsuit \: Let \: n \: \: be \: odd \: \: i.e. \: n = 2m + 1$

$\implies \sf{3x = (2m + 1)\pi} - \bigg\lgroup \sf{ \frac{π}{2} - x}\bigg\rgroup$

$\implies \sf{3x = 2m\pi + \frac{\pi}{2} + x } \\ \\ \implies \sf{2x = 2m\pi + \frac{\pi}{2} } \\ \\ \implies \sf{ \pink{x = m\pi + \frac{\pi}{4} }}....(ii)$

$\sf {Now, \: as \: - \frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}}$

$\dag \: \sf \bold{From \: Eqs. \: (i) \: and \: (ii)} \\ \\ \sf{ \therefore \: x = \frac{\pi}{8} ,\frac{\pi}{4}, - \frac{3\pi}{8} }$

$\sf{ \red{ \underline{Thus, \: \: point \: \: of \: \: intersection \: \: are}}}$

$\bigg\lgroup \sf{ \frac{\pi}{8}, \ \cos \frac{\pi}{8} }\bigg\rgroup,\bigg\lgroup \sf{ \frac{\pi}{4}, \ \cos \frac{\pi}{4} }\bigg\rgroup,\bigg\lgroup \sf{ - \frac{3\pi}{8}, \ \cos \frac{3\pi}{8} }\bigg\rgroup$

Answer 2

Answer:

\sf{ \clubs \: \: Let \: \: n \: \: be \: \: even \: \: i.e. \: n = 2m}♣Letnbeeveni.e.n=2m \\

\begin{gathered} \implies \sf{3x = 2m\pi + } \bigg\lgroup \\ \sf{ \frac{π}{2} - x}\bigg\rgroup\\\end{gathered}⟹3x=2mπ+⎩⎪⎪⎪⎧2π−x⎭⎪⎪⎪⎫ \\

\implies \sf {\pink{x = \frac{mn}{2} + \frac{\pi}{8} }} \: ....(i)⟹x=2mn+8π....(i)

\sf{} \clubsuit \: Let \: n \: \: be \: odd \: \: i.e. \: n = 2m + 1♣Letnbeoddi.e.n=2m+1

\begin{gathered} \implies \sf{3x = (2m + 1)\pi} - \bigg\lgroup \sf{ \frac{π}{2} - x}\bigg\rgroup\\\end{gathered}⟹3x=(2m+1)π−⎩⎪⎪⎪⎧2π−x⎭⎪⎪⎪⎫

\begin{gathered} \implies \sf{3x = 2m\pi + \frac{\pi}{2} + x } \\ \\ \implies \sf{2x = 2m\pi + \frac{\pi}{2} } \\ \\ \implies \sf{ \pink{x = m\pi + \frac{\pi}{4} }}....(ii)\end{gathered}⟹3x=2mπ+2π+x⟹2x=2mπ+2π⟹x=mπ+4π....(ii)

\sf {Now, \: as \: - \frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}}Now,as−2π⩽x⩽2π

\begin{gathered} \dag \: \sf \bold{From \: Eqs. \: (i) \: and \: (ii)} \\ \\ \sf{ \therefore \: x = \frac{\pi}{8} ,\frac{\pi}{4}, - \frac{3\pi}{8} }\end{gathered}†FromEqs.(i)and(ii)∴x=8π,4π,−83π

\sf{ \red{ \underline{Thus, \: \: point \: \: of \: \: intersection \: \: are}}}Thus,pointofintersectionare

\bigg\lgroup \sf{ \frac{\pi}{8}, \ \cos \frac{\pi}{8} }\bigg\rgroup,\bigg\lgroup \sf{ \frac{\pi}{4}, \ \cos \frac{\pi}{4} }\bigg\rgroup,\bigg\lgroup \sf{ - \frac{3\pi}{8}, \ \cos \frac{3\pi}{8} }\bigg\rgroup⎩⎪⎪⎪⎧8π, cos8π⎭⎪⎪⎪⎫,⎩⎪⎪⎪⎧4π, cos4π⎭⎪⎪⎪⎫,⎩⎪⎪⎪⎧−83π, cos83π⎭⎪⎪⎪⎫

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