# Trig Problem

• Sep 20th 2007, 11:20 AM
Tom G
Trig Problem
There are two similar questions in my homework which need me to solve equations and I have now got stuck on one of them and would like someone to tell me if I have made a mistake in my workings or what I should do next.

Q. Solve 2tan^2 x-7sec+8=0 for 0<x<360

So far I have got:

2tan^2 x-7sec+8=0

2(sin^2 x/cos^2 x)-7cosx+8=0

I am mainly confused because I want to multiply by cos^2 x in order to cancel out the cos^2 x under sin^2 x, but can I do this given that it is all multiplied by 2 i.e. in the brackets?:confused:

ALSO: I would be grateful if someone could clarify this for me.

Q. State the values of:

arc sin 0.5 - my answer is 30 degrees

arc tan 1 - my answer is 45 degrees

I ask this because I have no idea what the 'arc' means - does it affect the way I should answer the question as I have never encountered it before.

Many thanks in advance MHF!
• Sep 20th 2007, 12:00 PM
ticbol
2tan^2 x-7sec+8=0

2(sin^2 x/cos^2 x)-7cosx+8=0

That's okay except that it should be -7/cos x.

You wanted to multiply all by cos^2 x in order to clear the fraction? Very good! Always clear the fractions first.

Sure you can multiply all by cos^2 x even if "it is all multiplied by 2 i.e. in the brackets".

So, to continue,

2[sin^2(X) /cos^2(X)] -7/cosX +8 = 0
Multiply both sides by cos^2(X),
2sin^2(X) -7cosX +8cos^2(X) = 0
2[1 -cos^2(X)] -7cosX +8cos^2(X) = 0
2 -2cos^2(X) -7cosX +8cos^2(X) = 0
6cos^2(X) -7cosX +2 = 0
Factor that,
(3cosX -2)(2cosX -1) = 0

3cosX -2 = 0
cosX = 2/3
X = arccos(2/3) = 48.1896851 degrees, in the 1st quadrant.
Since cosine is positive also in the 4th quadrant,
X = 360 -48.1896851 = 311.8103149 deg, in the 4th quadrant.

2cosX -1 = 0
cosX = 1/2
X = arccos(1/2)) = 60, in the 1st quadrant.
Since cosine is positive also in the 4th quadrant,
X = 360 -60 = 300 deg, in the 4th quadrant.

Therefore, X = 48.1896851, 60, 300, or 311.8103149 degrees -----answer.

-----------------------------
ALSO: I would be grateful if someone could clarify this for me.

Q. State the values of:

arc sin 0.5 - my answer is 30 degrees

arc tan 1 - my answer is 45 degrees

I ask this because I have no idea what the 'arc' means - does it affect the way I should answer the question as I have never encountered it before.

arcsin(0.5) = 30 deg or 150 deg, since sine is positive in the 1st and 2nd quadrants.

arctan(1) = 45deg or 225deg, since tangent is positive in the 1st and 3rd quadrants.

'arc' here means, or, arcsin(0.5) means an anlge whose sine is 0.5.
arctan(1) is an angle whose tangent is 1.
So associate "arc___" with an angle.
• Sep 20th 2007, 12:01 PM
Soroban
Hello, Tom G!

Here's the first one . . .

Quote:

1) Solve: . $2\tan^2 x -7\sec x + 8\;=\;0$ . for $0^o \leq x \leq 360^o$

We have: . $2\left(\sec^2x - 1\right) - 7\sec x + 8 \;=\;0$

. . which simplifies to: . $2\sec^2x - 7\sec x + 6 \;=\;0$

. . which factors: . $(\sec x - 2)(2\sec x - 3) \;=\;0$

And has roots:

. . $\sec x - 2 \:=\:0\quad\Rightarrow\quad\sec x \:=\:2\quad\Rightarrow\quad x \:=\:60^o,\:300^o$

. . $2\sec x - 3 \:=\:0\quad\Rightarrow\quad\sec x \:=\:\frac{3}{2}\quad\Rightarrow\quad x \:\approx\:48.19^o,\:311.81^o$

• Sep 21st 2007, 05:40 AM
topsquark
Quote:

Originally Posted by Tom G
Q. State the values of:

arc sin 0.5 - my answer is 30 degrees

arc tan 1 - my answer is 45 degrees

I ask this because I have no idea what the 'arc' means - does it affect the way I should answer the question as I have never encountered it before.

The "arcsine" function is the "older" labeling of the inverse sine function. ie.
$sin(30^o) = \frac{1}{2} \implies arcsin \left ( \frac{1}{2} \right ) = sin^{-1} \left ( \frac{1}{2} \right) = 30^o$

You won't typically see the newer textbooks using asin, acs, atn, etc. (Please don't ask me for a cut-off date for this. For all I know there are some new texts using this.) For example I learned the "arc" functions in High School in the late 80's, but my Calc book in college didn't use them. However my Calc book that was published in the 70's does use them.

-Dan
• Sep 21st 2007, 09:06 AM
Tom G
Just one more question...

Does arc cos(-1/square root of 2) = -45 ?
• Sep 21st 2007, 09:18 AM
Jhevon
Quote:

Originally Posted by Tom G
Just one more question...

Does arc cos(-1/square root of 2) = -45 ?

no. (remember cosine is an even function)

couldn't you have plugged that in to your calculator? or you could work it out (it's actually not that hard)
• Sep 21st 2007, 11:40 AM
ticbol
Quote:

Originally Posted by Tom G
Just one more question...

Does arc cos(-1/square root of 2) = -45 ?

-45 degrees?
No.

-1/sqrt(2) is negative. (:))
In what quadrants is cosine negative?
In the 2nd and 3rd quadrants.
So, 135 deg and 225 deg. --------------answer.