Solving trigonometric functions Multiple choice HSC (2 Viewers)

Jihado31 · Jul 10, 2019

InteGrand said:
There should be two solutions for $\tan x + 2 = 0$ . You were meant to do 180º + (-63)º, and 360º + (-63)º for that.

So answer is that there are three solutions.

Alternatively, just sketch graphs of the two functions in the given domain. We can see then that the horizontal line $y=-2$ intersects that of $y=\tan x$ in two places in the given domain, and the line of $y=1$ touches the graph of $y=\sin x$ exactly once in the given domain. Hence answer is 3.

That is incorrect because when you said "You were meant to do 180º + (-63)º, and 360º + (-63)º for that." indicates that you think the forth quadrant is 360+theta. You were meant to write 360--63, which therefore means its outside the 0-360 range, thus making it 2 solutions. Though a question.. When you want to find theta, do you sub in shift tan -2 or 2??

InteGrand · Jul 20, 2019

Jihado31 said:
That is incorrect because when you said "You were meant to do 180º + (-63)º, and 360º + (-63)º for that." indicates that you think the forth quadrant is 360+theta. You were meant to write 360--63, which therefore means its outside the 0-360 range, thus making it 2 solutions. Though a question.. When you want to find theta, do you sub in shift tan -2 or 2??

$\noindent The solutions to $\tan x + 2 = 0$ for $x$ between $0^{\circ}$ and $360^{\circ}$ are indeed $180^{\circ} + \tan^{-1}(-2)$ and $360^{\circ} + \tan^{-1}(-2)$.$

$\noindent To see this, note that if we don't restrict $x$, then $\tan x + 2 = 0 \iff \tan x = -2$ has a solution $x = \tan^{-1}(-2)$ (which is something between $-90^{\circ}$ and $0^{\circ}$). Hence all possible solutions to the equation $\tan x + 2 = 0$ are found by adding on any multiple of $180^{\circ}$ to this, i.e. $x = \tan^{-1}(-2) + n\times 180^{\circ}$ for some integer $n$. Finally, the only values of $n$ that give us solutions between $0^{\circ}$ and $360^{\circ}$ are when $n = 1$ or $n = 2$ (because $\tan^{-1}(-2)$ is something between $-90^{\circ}$ and $0^{\circ}$). This is why I wrote what I wrote originally.$

$\noindent And to answer your last question, you would use $-2$, not $2$.$

Solving trigonometric functions Multiple choice HSC (2 Viewers)

Jihado31

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InteGrand

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