Question

$\large{{\underline{\color{red}{\bf{question - }}}}}$

$\large{{\underline{\color{green}{\bf{find \: the \: general \: solution \: for \: each \: in \: the \: equation - }}}}}$


➪ Sin2x + Cosx = 0


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Answer 1

Step-by-step explanation:

$\large\blue{ \underline{\underline{ \qquad\qquad \red{Given \: that} \qquad\qquad}}}$

• Sin2x + Cosx = 0

$\bf\large\blue{ \underline{\underline{ \qquad\qquad \red{Need\; To \: Find} \qquad\qquad}}} \:$

• find the general solution for each in the equation

$\rm \large\blue{ \underline{\underline{ \qquad\qquad \green{SoLuTiOn} \qquad\qquad}}} \:$

$\rm \: ➪ \: \: \: Sin2x + Cosx = 0 \: \: \: \qquad\{given \}$

Putting sin 2x = 2 sin x ços x

$\rm \blue{ : ➪ }\: \: \: \: 2 \: sin \: x \: cos \: x \: + cos \: x \: = 0 \\\\\\ \large \rm \blue{: \: ➪ \: \: \:cos \: x \: (2sin \: x + 1) = 0 \: }$

Hence ,

$\tt❒ \: cos \: x = 0 \: \: \: \& \: \: \: 2sin \: x + 1 = 0 \\\\\\ \tt❒cos \: x \: = 0 \: \: \: \& \: \: \: 2sin \: x = - 1 \\\\\\ \large {\tt❒ \: \pink{cos \: x \: = 0 \: \: \: \& \: \: \: sin \: x \dfrac{ - 1}{2} }}$

■we find general solution of both equations separately

❒ cos x = 0 (Given)

$\sf \: general \: solution \: be \: \: \red{x \: = (2n + 1) \dfrac{ \pi}{2} } \qquad \:$

$\tt \qquad \qquad \sf \: \: \: where \bf \: \: n \: \in \: \: z$

$\sf \: General \: solution \: for \: \red{ sin \: x = \dfrac{ - 1}{2} }$

$\sf \: ❒ Let \red{\: s in \: x = sin \: y } \qquad \: - - - eq.(1)$

$\sf❒ \red{ sin \: x = \dfrac{-1}{2} } \qquad - - - (1) \qquad \bigg(given \bigg)$

_____________________________

$\huge \qquad \qquad \frak{rough}$

we know that

$\sf \: sin \: 30 \degree \: = \dfrac{1}{2}$

$\qquad \sf \: But \: we \: need \: \dfrac{ - 1}{2}$

$\bf \: So, \: angle \: is \: 3^{ rd} \: \& \: 4^ {th} \: quadrant \: \: \theta = 30 \degree$

$:\longrightarrow \sf \: 180 + \theta \\ \\ \\ :\longrightarrow180 + (30) \\ \\ \\ \large \pink{ :\longrightarrow210 \degree}$

$:\longrightarrow \sf \: \cancel{210} \times \dfrac{\pi}{ \cancel{180} } \\ \\ \\ :\longrightarrow7 \times \dfrac{\pi}{6} \\ \\ \\ \: \pink{ : \large \longrightarrow \dfrac{7\pi}{6} }$

______________________________

♤From equation (1) & (2)

$\sf: \longmapsto \: sin \: x = sin \: y$

substituting values we get

$\sf: \longmapsto \: sin \: y \: = \: \dfrac{ - 1}{2}$

$\sf: \longmapsto \cancel{ sin }\: y = \cancel{sin }\: \dfrac{7 \pi}{6} \\\\\large\sf y = \dfrac{7\pi}{6}$

General Solution is

$\sf \: x \: = n \pi \: + ( - 1)^ny \: \: \: where \: \bf \: n \: \in \: z$

putting value of y = $\dfrac{7\pi}{6}$

we get,

$\sf \: x \: = n \pi \: + ( - 1) ^n \dfrac{7 \pi}{6} \: \: \: \: \: where \bf \: n \: \in \: z$

Therefore

$\sf \: \: for \: cos \: x \: = 0 \: , \: \: x = (2n + 1) \dfrac{ \pi}{2} \\ \\ \\ \bf \large \: or$

$\sf for \: sin \: x \: = \dfrac{ - 1}{2} \: \: \: , \: \: x =n \pi \: + ( - 1)^n \: \dfrac{7 \pi}{6} \: \: \: \: where \: \bf \: n \: \in \: x$