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Is this the first time you saw this question? If so then I understand why this question felt a bit difficult. Here let us start. Step 1: \int{\sin^{2}x\cos^{3}xdx}=\int{\sin^{2}x\cos^{2}x\cdot{\cos{x}}dx} Step 2...

Good attempt however this is beyond the scope of the NSW Mathematics Extension I syllabus.

It should be less than \frac{\pi}{2}+k\pi but greater than or equal to \frac{\pi}{4}+k\pi. The other solution is because when x=\frac{\pi}{2} the result is undefined or infinity so therefore, the value has to be smaller than x=\frac{\pi}{2} and the +k\pi comes from the fact the tan function...

Simple play \sin{2x}-2\sin^{2}x>0 2\sin{x}\cos{x}-2\sin^{2}x>0 2\sin{x}\left(\cos{x}-\sin{x}\right)>0 Now we use something called the auxiliary angle for \left(\cos{x}-\sin{x}\right)=R\cos\left(x+\alpha\right) Expand\;R\cos\left(x+\alpha\right)=R\cos{x}\cos{\alpha}-R\sin{x}\sin{\alpha}...

I have a different take on this one, given the pattern of the binomial question you will have \left(1-x\right)^{n}=1-\begin{pmatrix}n\\1\end{pmatrix}x+\begin{pmatrix}n\\2\end{pmatrix}x^{2}+...+\begin{pmatrix}n\\n\end{pmatrix}\left(-x\right)^{n} Integrate that and you will have...

The problem is that \cos{\frac{\pi}{3}}+|\sin\frac{\pi}{6}| is actually 1 which is 1\neq{0}

\frac{1}{\sqrt{1+\cot^{2}x}}and that to help you understand here is a simple example \sqrt{x^{2}}=\pm{x}=|x|

i re-did the question based off of this and ended up getting the appropriate solutions, it didn't seem obvious to me before, but rather than simplifying 1 + cot^2(x) to csc^2(x), i changed cot to cos/sin, then with some simplifying you end up with cos2x + |sinx| = 0, then you solve considering...

The problem is that \cos{\frac{\pi}{3}}+|\sin\frac{\pi}{6}| is actually 1 which is 1\neq{0}

You can actually rule out the \pm\frac{\pi}{6}+k\pi solution because the equation does not match.

I believe that instead of \cos{2x}+\sin{x} the question is \cos{2x}-\sin{x} and next step you just have 1-2\sin^{2}x-\sin{x}=0 2\sin^{2}x+\sin{x}-1=0 \left(\sin{x}+1\right)\left(2\sin{x}-1\right) \sin{x}=-1, \frac{1}{2} \sin{x}=\pm\frac{\pi}{2}+2k\pi, \pm\frac{\pi}{6}+k\pi. This I believe is...

With the \frac{-\pi}{2}+2k\pi solution I believe that they took the negative root as well which can be justified through similar working.

When will people ever learn that some of the old questions are still related to the new syllabus like inverse functions.

With part b is the volume just a donut and from memory the volume is simply a circle multiplied by the dough formed by the circumference of the circle being rotated around the y-axis.

y=\frac{4x^{2}}{x^{2}-9}=4+\frac{36}{x^{2}-9} First, off focus on \frac{36}{x^{2}-9} differentiate that we will have \frac{-2x}{\left(x^{2}-9\right)^{2}} and if you differentiate it again then we will have...

That would be correct if the question was find three cube roots of the complex number 8i but instead the answer is \frac{3\sqrt{3}}{2\sqrt[3]{2}}

Step 1 Draw an Argand diagram with the complex number specified Step 2 Write in mod-arg form. \sqrt{3}+i=2\left(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\right) Step 3 Rewrite step 2 but instead modify the mod-arg form so that it looks like this...

Leave it as \tan{x} and draw a triangle to make your life easier.

That tells us something. Homework\;too\;difficult\rightarrow{Student\;asks\;Dr\;Du\;tutor\;problematic\;maths\;question}\rightarrow{Dr\;Du\;tutor\;says\;look\;at\;the\;examples}\rightarrow{Student...

If I want to write something then I can say people who go to Dr Du, keep going there because it gives us free maths questions to look at due to continuous anecdotes from students that the Dr du centre does not provide effective help for its students and tells students to look at the notes given...