The Last Few Problems of my Summer Assignment D:

[evenkeel3x]

So I've completed 100+ problems for this enormous AP Calc summer assignment, and the only problems left are some really difficult review ones from math analysis. I posted this under Geometry and Trig, because most of the confusing ones are Trig-Related. So, rest assured I have attempted all of these repeatedly, to no avail, I get one or two steps done and find out they're pretty much useless, leading down a dead end. So I'm not going to include all the steps, because they'd be all wrong anyways. These are the last few problems, so I'd really appreciate all the help I can get on them.

the first one:
Find all the solutions:
2sin^3X + sin^2X = 0

Next:
Prove the Identity:
1 + tan^2X =csc^2X
tan^2X
I hope that turns out typed right...

Simplify:
(sinX) (sinX) + (cosX) (cosX)
(secX)

Prove the identity:
sinX + sinXcot^2X = cscX

And last but not least...
Simplify:
2X^2e^X - 6Xe^X
(2e^X)^2

These are the last few problems on this rediculously long assignment, I would appreciate any help I could get from you guys, as I am really stuck. Thanks!

 

[soroban]

Hello, evenkeel3x!

Good work on the other 100+ problems!

But I must comment that these problems are quite straight-forward
. . and I wonder where your difficulties lie.

\(\displaystyle \text{Find all solutions: }\;2\sin^3\!x + \sin^2\!x \:=\: 0\)

\(\displaystyle \text{Factor: }\:\sin^2\!x(2\sin x + 1) \:=\:0\)

\(\displaystyle \text{And we have two equations to solve:}\)

. . \(\displaystyle \sin^2\!x \:=\:0 \quad\Rightarrow\quad \sin x \:=\:0 \quad\Rightarrow\quad x \:=\:\pi n\)

. . \(\displaystyle \2\sin x + 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:-\frac{1}{2} \quad\Rightarrow\quad x \:=\:\begin{Bmatrix}\frac{7\pi}{6} + 2\pi n \\ \text{-}\frac{\pi}{6} + 2\pi n \end{Bmatrix}\)

\(\displaystyle \text{Prove the Identity: }\:\frac{1 + \tan^2\!x}{\tan^2\!x} \:=\:\csc^2\!x\)

\(\displaystyle \frac{1+\tan^2\!x}{\tan^2\!x} \;=\;\frac{1}{\tan^2\!x} + \frac{\tan^2\!x}{\tan^2\!x} \;=\;\cot^2\!x + 1 \;=\;\csc^2\!x\)

\(\displaystyle \text{Simplify: }\: \frac{(\sin x) (\sin x) + (\cos x) (\cos x)}{\sec x}\)

\(\displaystyle \frac{(\sin x)(\sin x) + (\cos x)(\cos x)}{\sec x} \;=\;\frac{\sin^2\!x + \cos^2\!x}{\sec x} \;=\; \frac{1}{\frac{1}{\cos x}} \;=\;\cos x\)

\(\displaystyle \text{Prove the identity: }\:\sin x + \sin x\cot^2\!x \:=\: \csc x\)

\(\displaystyle \text{The left side is: }\;\sin x(1 + \cot^2\!x) \;=\;\sin x\cdot\csc^2x \;=\;\frac{1}{\csc x}\cdot\csc^2\!x \;=\;\csc x\)

\(\displaystyle \text{Simplify: }\: \frac{2x^2e^x - 6xe^x}{(2e^x)^2}\)

\(\displaystyle \text{We have: }\;\frac{2xe^x(x-3)}{4e^{2x}} \;=\; \frac{x(x-3)}{2e^x}\)

 

[evenkeel3x]

Thanks very much, I'm glad you actually helped unlike other users on this website who critique grammar more than math... I think the problem for me is the trig, the identities just don't seem to stick!
Thanks again!