anonymous
  • anonymous
Solve without a calculator
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\[\huge 4\tan^2-\sec^2x=0\]\]
anonymous
  • anonymous
0
anonymous
  • anonymous

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UnkleRhaukus
  • UnkleRhaukus
\[4\tan^2 (x)-\sec^2 (x)=0\]
anonymous
  • anonymous
yup
UnkleRhaukus
  • UnkleRhaukus
\[\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\]\[\sec(\theta)=\frac{1}{\cos(\theta)}\]
UnkleRhaukus
  • UnkleRhaukus
•take the sec term to the other side of the equation, •substitute the above definitions into your equation, •multiply by the denominator •divide by 4 •take the sqrt of both sides , •take the inverse trig function to isolate x
UnkleRhaukus
  • UnkleRhaukus
tell me if you get stuck.
anonymous
  • anonymous
\[4\frac{ \sin^2x }{ \cos^2x }-\frac{ 1 }{ \cos^2x }=0\]
anonymous
  • anonymous
what do i do next
UnkleRhaukus
  • UnkleRhaukus
multiply by cos^2
anonymous
  • anonymous
to subtract the two ??
anonymous
  • anonymous
(2sinx+1)(2sinx-1) / cos^2x
UnkleRhaukus
  • UnkleRhaukus
how did that happen?
anonymous
  • anonymous
idk im guessing :S
UnkleRhaukus
  • UnkleRhaukus
start again and follow the •'s
anonymous
  • anonymous
okay
anonymous
  • anonymous
i'm going to be honest, im confused
anonymous
  • anonymous
by following the checlist this is what im getting sinxcosx-sinx
Hero
  • Hero
May I suggest another approach?
UnkleRhaukus
  • UnkleRhaukus
...
anonymous
  • anonymous
my teacher has a completely different way im confuseeed
UnkleRhaukus
  • UnkleRhaukus
well im sure there are many different ways ,
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
1 + tan^2 x = sec^2 x tan^2 x = sec^2 x - 1
UnkleRhaukus
  • UnkleRhaukus
i dont like that way as much as i like the way i have outlined , but that is opinion
anonymous
  • anonymous
4 ( sec^2 x - 1) - sec^2 x = 0 4 sec^2 x - 4 - sec^2 x = 0 3 sec^2 x = 4 sec x = 2/ sqrt (3)
anonymous
  • anonymous
i am only allowed to have cos and sin in my answers
anonymous
  • anonymous
Lol.....u shuld have Mentioned that in the question...)
anonymous
  • anonymous
Steps: 1. rearrange and isolate the unknown 2. move everything to one since, set equal to zero 3. use the zero principle 4. use inverse trig to find the angle 5. use CAST or ASTC to adjust answers @Yahoo! i know sorry ! :$
Hero
  • Hero
|dw:1355392376220:dw|
anonymous
  • anonymous
guys I GOT IT ! :D
anonymous
  • anonymous
i just used my teachers method but everyone's explanations helped. thankyou :D
Hero
  • Hero
And what exactly was your teacher's method?
anonymous
  • anonymous
we're supposed to find what x can and cannot equal
anonymous
  • anonymous
so i just used basic trig and then looking at the unit circle i did elimination and listed the factors
Hero
  • Hero
That's not what you posted at the beginning
anonymous
  • anonymous
burhan101 0 Steps: 1. rearrange and isolate the unknown 2. move everything to one since, set equal to zero 3. use the zero principle 4. use inverse trig to find the angle 5. use CAST or ASTC to adjust answers
anonymous
  • anonymous
thats what i did, using the zero principle :
Hero
  • Hero
You mind showing some of those steps using the drawing button like I did?
anonymous
  • anonymous
|dw:1355393054138:dw|
Hero
  • Hero
Kill your teacher bro
UnkleRhaukus
  • UnkleRhaukus
\[4\tan^2 (x)-\sec^2 (x)=0\]\[4\tan^2 (x)=\sec^2 (x)\]\[4\left(\frac{\sin(x)}{\cos (x)}\right)^2 =\left(\frac1{\cos(x)}\right)^2\]\[4\frac{\sin^2(x)}{\cos^2 (x)} =\left(\frac1{\cos^2(x)}\right)\]\[4\sin^2(x)=\frac{\cos^2(x)}{\cos^2(x)}\]\[4\sin^2(x) =1\]\[\sin^2(x) =\frac14\]\[\sin(x)=\frac 12\]\[x=\arcsin\left(\frac12\right)=60°\]
Hero
  • Hero
@UnkleRhaukus, you ignore negative values?
UsukiDoll
  • UsukiDoll
wow Hero that's just harsh, but...yeah..no comment
UsukiDoll
  • UsukiDoll
I'm not saying it sucks or anything though
anonymous
  • anonymous
\[4\tan^2 (x)-\sec^2 (x)=0\]\[4\tan^2 (x)-(1+tan^2x)=0\]\[3\tan^2 (x)-1=0\]\[tan^2x = \frac{1}{3}\]That shouldn't be difficult.
anonymous
  • anonymous
Just use the identity \(1+tan^2x=sec^2x\)
UnkleRhaukus
  • UnkleRhaukus
Hero
  • Hero
I didn't see that second line. He posted it in pieces.

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