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\[\[\huge 4\tan^2-\sec^2x=0\]\]
0

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Other answers:

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

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