Weird MCQ options!

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February 5th 2010, 06:32 AM
Punch

Weird MCQ options!

Determine all the values of angle $x$ if $3sinx=tanx$ and $x$ is between $0^\circ$ and $360^\circ$ . Give the answers correct to 1 decimal place.

$[A] 70.5^\circ, 180^\circ, 250.5^\circ$

$[B] 0^\circ, 70.5^\circ, 180^\circ, 289.5^\circ, 360^\circ$

$[C] 0^\circ, 70.5^\circ, 180^\circ, 250.5^\circ, 360^\circ$

$[D] 70.5^\circ, 180^\circ, 289.5^\circ$

This is weird, I have solved to the point that $cosx=\frac{1}{3}$ and $x$ is in $Q_1, Q_4$ . Hence one of the options should have only 2 answers. Did I go wrong?
February 5th 2010, 06:47 AM
e^(i*pi)

Quote:

Originally Posted by Punch

Determine all the values of angle $x$ if $3sinx=tanx$ and $x$ is between $0^\circ$ and $360^\circ$ . Give the answers correct to 1 decimal place.

$[A] 70.5^\circ, 180^\circ, 250.5^\circ$

$[B] 0^\circ, 70.5^\circ, 180^\circ, 289.5^\circ, 360^\circ$

$[C] 0^\circ, 70.5^\circ, 180^\circ, 250.5^\circ, 360^\circ$

$[D] 70.5^\circ, 180^\circ, 289.5^\circ$

This is weird, I have solved to the point that $cosx=\frac{1}{3}$ and $x$ is in $Q_1, Q_4$ . Hence one of the options should have only 2 answers. Did I go wrong?

You didn't go wrong but rather didn't complete the whole equation. The problem you made appears to be that you cancelled out sin(x) which you can't do because sin(x) is potentially 0. I've outlined the method using factorising below

From the domain restrictions $x \neq 90^{\circ} + 180k \: \: , \: k \in \mathbb{Z}$

$3\sin(x) - \frac{\sin(x)}{\cos(x)} = 0$

$\frac{3\sin(x) \cos(x) - \sin(x)}{cos(x)}$

Multiply by cos(x) and factorise

$\sin(x) (3\cos(x)-1) = 0$

$\sin(x) = 0$ or $\cos(x) = \frac{1}{3}$

$x = 0\: ,\: 180\: ,\: 360$

$x = 70.5\: , \: 289.5$

This is because cos(x) is symmetrical in the line $x = 180^{\circ}$ so we can do $180+(180-70.5)$ where 180-70.5 is the distance the first point is away from 180
February 5th 2010, 06:55 AM
Punch

Quote:

Originally Posted by e^(i*pi)

You didn't go wrong but rather didn't complete the whole equation. The problem you made appears to be that you cancelled out sin(x) which you can't do because sin(x) is potentially 0. I've outlined the method using factorising below

From the domain restrictions $x \neq 90^{\circ} + 180k \: \: , \: k \in \mathbb{Z}$

$3\sin(x) - \frac{\sin(x)}{\cos(x)} = 0$

$\frac{3\sin(x) \cos(x) - \sin(x)}{cos(x)}$

Multiply by cos(x) and factorise

$\sin(x) (3\cos(x)-1) = 0$

$\sin(x) = 0$ or $\cos(x) = \frac{1}{3}$

$x = 0\: ,\: 180\: ,\: 360$

$x = 70.5\: , \: 289.5$

This is because cos(x) is symmetrical in the line $x = 180^{\circ}$ so we can do $180+(180-70.5)$ where 180-70.5 is the distance the first point is away from 180

So i Cant cancel sinx
February 5th 2010, 07:08 AM
e^(i*pi)

Quote:

Originally Posted by Punch

So i Cant cancel sinx

Nope. Attached is the graph of this question, on it you can see that 0 and 180 are definitely crossing points.

If you'd cancelled sin(x) you'd have been dividing by 0 at the points 0, 180 and 360.

In general it's always better to look for a factorising solution first
February 5th 2010, 09:54 PM
Punch

I will move everything to the left side or right side before factorizing the common items and equating them to values before solving... thanks!