SOLUTION: Hi, can you please help me with this question Find the range of x in the interval [0, 2π) for which 8 sin x + cos x > 4. The answers are apparently 2arctan(0.2) and 2 arctan(3).

Algebra ->  Trigonometry-basics -> SOLUTION: Hi, can you please help me with this question Find the range of x in the interval [0, 2π) for which 8 sin x + cos x > 4. The answers are apparently 2arctan(0.2) and 2 arctan(3).      Log On


   



Question 1203483: Hi, can you please help me with this question
Find the range of x in the interval [0, 2π) for which 8 sin x + cos x > 4.
The answers are apparently 2arctan(0.2) and 2 arctan(3). I need help with my explanation. Thank you in advance.

Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
In your words
"Find the range of x in the interval [0, 2π) for which 8 sin x + cos x > 4."
I think the word "range" should be "domain". Anyway:

8sin%28x%29+%2B+cos%28x%29+%3E+4%29, 0%3C=x%3C2pi

matrix%281%2C4%2CLet%2Cx=2y%2Cor%2Cy=x%2F2%29

Then since 0%3C=x%3C2pi, 0%3C=2y%3C2pi or 0%3C=y%3Cpi

8sin%282y%29+%2B+cos%282y%29+%3E+4%29, 0%3C=y%3Cpi

The terms on the left are trig expressions. To make the term 4 on 
the right become a trig expression also, we play the clever trick 
of multiplying the 4 by 1, then replacing the 1 by the familiar
Pythagorean trig identity 1=cos%5E2%28y%29%2Bsin%5E2%28y%29:

We use the formulas for sine and cosine of 2y:

8%282sin%28y%29cos%28y%29%29%2B%28cos%5E2%28y%29-sin%5E2%28y%29%29+%3E+4%281%29



16sin%28y%29cos%28y%29%2Bcos%5E2%28y%29-sin%5E2%28y%29+%3E+4cos%5E2%28y%29%2B4sin%5E2%28y%29

16sin%28y%29cos%28y%29%2Bcos%5E2%28y%29-sin%5E2%28y%29+%3E+4cos%5E2%28y%29%2B4sin%5E2%28y%29

16sin%28y%29cos%28y%29-3cos%5E2%28y%29-5sin%5E2%28y%29+%3E+0

-5sin%5E2%28y%29%2B16sin%28y%29cos%28y%29-3cos%5E2%28y%29+%3E+0

5sin%5E2%28y%29-16sin%28y%29cos%28y%29%2B3cos%5E2%28y%29+%3C+0

%285sin%28y%29-cos%28y%29%29%28sin%28y%29-3cos%28y%29%29%3C0

One of those factors is positive and the other is negative. So either

system%285sin%28y%29-cos%28y%29%3E0%2CAND%2Csin%28y%29-3cos%28y%29%3C0%29 or system%285sin%28y%29-cos%28y%29%3C0%2CAND%2Csin%28y%29-3cos%28y%29%3E0%29

Those would involve only tan(y) if we divided through by cos(y). We could 
do that as long as cos(y) is not 0.  Let's make sure cos(y) is not zero.

If cos(y) were 0, then y=pi%2F2 so let's substitute y=pi%2F2 in
%285sin%28y%29-cos%28y%29%29%28sin%28y%29-3cos%28y%29%29%3C0
%285sin%28pi%2F2%29-cos%28pi%2F2%29%29%28sin%28pi%2F2%29-3cos%28pi%2F2%29%29%3C0
%285%281%29-0%29%281-3%280%29%29%3C0
5%3C0<--false!

So cos(y) can't be 0. Therefore, we can divide by it. For the first case

system%285sin%28y%29-cos%28y%29%3E0%2CAND%2Csin%28y%29-3cos%28y%29%3C0%29



system%285%2Atan%28y%29-1%3E0%2CAND%2Ctan%28y%29-3%3C0%29

system%28tan%28y%29%3E1%2F5=0.2%2CAND%2Ctan%28y%29%3C3%29

Since tangent is a strictly increasing function,

system%28y+%3E+arctan%280.2%29%2C+AND%2Cy%3Carctan%283%29%29

arctan%280.2%29+%3C+y+%3C+arctan%283%29

Now since y=x%2F2,

arctan%280.2%29+%3C+x%2F2+%3C+arctan%283%29

2arctan%280.2%29+%3C+x+%3C+2arctan%283%29

That's the interval you were looking for.  If we considered the other case

system%285sin%28y%29-cos%28y%29%3C0%2CAND%2Csin%28y%29-3cos%28y%29%3E0%29

it would lead to

system%28tan%28y%29%3C1%2F5=0.2%2CAND%2Ctan%28y%29%3E3%29

which is a contradiction, so the domain is

2arctan%280.2%29+%3C+x+%3C+2arctan%283%29

Edwin

Answer by ikleyn(52817) About Me  (Show Source):