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Could someone point me in the right direction? Thanks!

 

If x is an angle such that tan x + sec x = 2, then evaluate cos x.

 Oct 16, 2022
 #1
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evaluate cos x

 

Hello Guest!

 

\(tan\ x + sec\ x = 2\\ sec\ x=1/cos\ x\\ tan\ x=\dfrac{\sqrt{1-cos^2x}}{cos\ x}\\ \dfrac{\sqrt{1-cos^2x}}{cos\ x}+\dfrac{1}{cos\ x}=2\\ \sqrt{1-cos^2x}+1=2cos\ x \)

\(\sqrt{1-cos^2x}=2cos\ x -1\ |\ cos\ x=y\\ 1-y^2=(2y-1)^2\\ 1-y^2=4y^2-4y+1\\ 5y^2-4y=0\\ y\in \{0,0.8\}\ |\ y=cos\ x\\ \color{blue}cos\ x\in \{0,0.8\}\)

laugh  !

 Oct 16, 2022
edited by asinus  Oct 16, 2022
edited by asinus  Oct 16, 2022
 #2
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+1

In answer #1, if \(\cos x = 0,\)  then \(\sec x\)  will be infinite and so should be rejected.

Here are two alternative routines.

\(\tan x + \sec x = 2, \\ \sec x =2 - \tan x, \\ \sec^{2}x=(2 - \tan x)^{2}, \\ 1+ \tan^{2}x = 4-4\tan x+\tan^{2}x, \\ 4 \tan x = 3, \\ \tan x = 3/4. \)

The 3 and the 4 can be seen as two legs of a 3, 4, 5 rt-angled triangle from which we can read off the other ratios.

\(\sin x = 3/5, \cos x = 4/5,\sec x = 5/4,\)

and there are two others.

Note that \(\tan x + \sec x = 3/4+5/4 = 8/4 =2.\)

Alternatively,

\(\tan x + \sec x = \frac {\sin x}{ \cos x }+\frac{1}{\cos x} = 2,\\ \sin x + 1 = 2\cos x, \\ (\sin x+1)^{2}=4\cos^{2}x,\\ \sin^{2}x+2\sin x + 1 = 4(1-\sin^{2} x ),\\ 5\sin^{2}x+2\sin x -3=0,\\ (5\sin x-3)(\sin x +1)=0, \)

\(5\sin^{2}x +2\sin x -3 = 0, \\(5\sin x -3)(\sin x +1)=0,\\ \sin x = 3/5,\)

etc.

(the second root sin x = -1 is rejected because that implies that cos x = 0 meaning sec x is infinite.)

 Oct 17, 2022

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