Question

Solve the trigonometric equation given by $\sf\frac\ \textless \ br /\ \textgreater \ { (sin(2x)-cos(x))}{cos(2x)+sin(x)-1)}$ $\sf{ = \: 0 \: for \: \: ≤ \: x \: ≤ \: 2pi }$
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Answer 1

Explanation:

f ( x ) / g ( x ) = 0 ,

So , f ( x ) = 0

In above question f ( x ) is sin ( 2 x ) - cos (x )

sin ( 2 x ) - cos (x) = 0

We know that sin ( 2 x ) = 2 sin x cos x

So , 2sinx cos x - cos ( x) = 0

Take cos x common

So, cos x ( 2 sin x - 1 ) = 0

Then , cos x = 0 , and 2 sin x - 1 = 0

First , cos x = 0

So , cos x = cos π / 2 = 0 ( x = π/2 )

cos x = cos 3 π / 2 = 0 ( x = 3 π / 2 )

Now , 2 sin x - 1 = 0

2 sin x - 1 = 2 sin π / 6 - 1 = 0 ( x = π/6 )

Again ,

2 sin x - 1 = 2 sin 5π / 6 - 1 = 0 ( x = 5π/6 )

Since equation is undefined for π / 6 and 5 π / 6

So , x = π / 2 and x = 3 π / 2

Answer 2

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

Given that

$\rm :\longmapsto\:0 \leqslant x \leqslant 2\pi$

and

$\rm :\longmapsto\:\dfrac{sin2x - cosx}{cos2x + sinx - 1} = 0$

We know

$\purple{ \boxed{ \bf{sin2x \: = \: 2 \: sinx \: cosx \: }}}$

and

$\purple{ \boxed{ \bf{ \: \: cos2x \: = \: 1 \: - \: {2sin}^{2}x \: \: }}}$

On substituting the identity of sin2x in numerator and cos2x in denominator, we get

$\rm :\longmapsto\:\dfrac{2sinxcosx - cosx}{1 - 2 {sin}^{2}x + sinx - 1 } = 0$

$\rm :\longmapsto\:\dfrac{2sinxcosx - cosx}{ \cancel1 - 2 {sin}^{2}x + sinx - \cancel1 } = 0$

$\rm :\longmapsto\:\dfrac{2sinxcosx - cosx}{ - 2 {sin}^{2}x + sinx } = 0$

$\rm :\longmapsto\:\dfrac{cosx \: \: \cancel{(2sinx - 1)}}{ - sinx \: \: \cancel{(2sinx - 1)}} = 0$

$\rm :\longmapsto\: - \: cotx = 0$

$\rm :\longmapsto\:cotx = 0$

$\bf\implies \:x = \dfrac{\pi}{2} \: \: or \: \: \dfrac{3\pi}{2}$

Verification :-

Given equation is

$\rm :\longmapsto\:\dfrac{sin2x - cosx}{cos2x + sinx - 1} = 0$

$\red{\rm :\longmapsto\:When \: x = \dfrac{\pi}{2}}$

On substituting the value in above equation, we get

$\rm :\longmapsto\:\dfrac{sin\pi - cos\dfrac{\pi}{2}}{cos\pi + sin\dfrac{\pi}{2} - 1} = 0$

$\rm :\longmapsto\:\dfrac{0 - 0}{ - 1 + 1 - 1} = 0$

$\rm :\longmapsto\:\dfrac{0}{ - 1} = 0$

$\bf\implies \:0 = 0$

Hence, verified

Now,

Again, Given equation is

$\rm :\longmapsto\:\dfrac{sin2x - cosx}{cos2x + sinx - 1} = 0$

$\red{\rm :\longmapsto\:When \: x = \dfrac{3\pi}{2}}$

$\rm :\longmapsto\:\dfrac{sin3\pi - cos\dfrac{3\pi}{2}}{cos3\pi + sin\dfrac{3\pi}{2} - 1} = 0$

$\rm :\longmapsto\:\dfrac{0 - 0}{ - 1 + 1 - 1} = 0$

$\rm :\longmapsto\:\dfrac{0}{ - 1} = 0$

$\bf\implies \:0 = 0$

Hence, Verified

Additional Information :-

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