Question

Find the general solution of the following equation cosx=-root3/2​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{cos(x)=-\dfrac{\sqrt{3}}{2}}$

$\tt{\implies\,cos(x)=-cos\left(\dfrac{\pi}{6}\right)}$

$\tt{\implies\,cos(x)=cos\left(\pi-\dfrac{\pi}{6}\right)}$

$\tt{\implies\,cos(x)=cos\left(\dfrac{5\pi}{6}\right)}$

$\tt{\implies\,x=2n\pi\,\pm\,\dfrac{5\pi}{6}}$

Answer 2

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

Given Trigonometric equation is

$\rm :\longmapsto\:cosx \: = \: - \: \dfrac{ \sqrt{3} }{2}$

can be re - written as

$\rm :\longmapsto\:cosx = \: - \: cos\bigg[\dfrac{\pi}{6} \bigg]$

We know,

$\boxed{ \tt{ \: cos(\pi - x) \: = - \: cosx \: }}$

So, using this, we get

$\rm :\longmapsto\:cosx = cos\bigg[\pi - \dfrac{\pi}{6} \bigg]$

$\rm :\longmapsto\:cosx = cos\bigg[ \dfrac{5\pi}{6} \bigg]$

We know,

$\boxed{ \tt{ \: cosx = cosy\rm \implies\: \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z \: }}$

So, using this, we get

$\rm \implies\: \boxed{ \tt{ \: x = 2n\pi \pm \: \dfrac{5\pi}{6} \: \forall \: n \in \: Z \: }}$

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More to know :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}$