Weird MCQ options!

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

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

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

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

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

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

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Originally Posted by Punch

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

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

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

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

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

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

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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 $\displaystyle x \neq 90^{\circ} + 180k \: \: , \: k \in \mathbb{Z}$

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

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

Multiply by cos(x) and factorise

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

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

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

$\displaystyle x = 70.5\: , \: 289.5$

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

Quote:

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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 $\displaystyle x \neq 90^{\circ} + 180k \: \: , \: k \in \mathbb{Z}$

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

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

Multiply by cos(x) and factorise

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

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

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

$\displaystyle x = 70.5\: , \: 289.5$

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

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So i Cant cancel sinx

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Originally Posted by Punch

So i Cant cancel sinx

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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

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