Question

Determine the general solution of 2 sin x.cosx- cos x = 0​

Answer 1

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

➢ Given Trigonometric equation is

$\rm :\longmapsto\:2sinx \: cosx - cosx= 0$

$\rm :\longmapsto\:cosx(2sinx - 1) = 0$

$\rm :\implies\:cosx = 0 \: \: \: or \: \: \: 2sinx - 1 = 0$

Consider,

$\rm :\longmapsto\:\:cosx = 0$

$\bf\implies \:x = (2n + 1)\dfrac{\pi}{2} \: \: where \: n \: \in \: Z$

Now,

Consider,

$\rm :\longmapsto\:2sinx - 1 = 0$

$\rm :\longmapsto\:2sinx = 1$

$\rm :\longmapsto\:sinx = \dfrac{1}{2}$

$\rm :\longmapsto\:sinx =sin \dfrac{\pi}{6}$

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

Additional Information :-

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

$\: \: \: \: \: \: \: \bull \bf \: \: \: \: \: \: where \: n \: \in \: Z$

Lets solved some more examples!!!

Find the principal solution of the following:-

1. 2sinx = 1

can be rewritten as sinx = 1/2

Since, sinx is positive in Ist and 2nd quadrant.

So, x = 30° or 180° - 30°

i.e. x = 30° or 150°.

2. 2cosx + 1 = 0

can be rewritten as cosx = - 1/2

Since, cosx is negative in 3rd and 4th quadrant.

So, x = 180° - 60° or 180° + 60°

i.e. x = 120° or 240°.