Question

Solve the equation in the interval ( 0, 2pi )
tanx + secx= Sqrt 3

Answer 1

Recall:
tan^2(x) + 1 = sec^2(x)

Answer 2

please note that your equation can be rewritten as below:
\[\frac{ 1+\sin x }{ cos x }=\sqrt{3}\]

Answer 3

I did get that far but I don't know what to do after that

Answer 4

so:
\[1+\sin x=\sqrt{3}\cos x\]
then:
\[(1+\sin x)^{2}=3(\cos x)^{2}\]
\[(1+\sin x)^{2}=3*(1-\sin x)(1+\sin x)\]
finally:
\[(1+\sin x)[[1+\sin x-3(1-\cos x)]=0\]

Answer 5

oh ok

Answer 6

please semplify the term in square bracket, furthermore I used the fundamental identitu:
cos^2+sin^2=1

Answer 7

oops: ...identity...

Answer 8

oops sorry:
\[(1+\sin x)[1+\sin x-3(1-\sin x)]=0\]

Answer 9

hmm you don't need to square
\(\sqrt{3}\cos x -\sin x =1 =====> \frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x=\frac{1}{2}\)

Answer 10

how did you replace cosx^2 with (1-sinx)(1+sinx)

Answer 11

that's something we know already

Answer 12

yes, I did!

Answer 13

xapproachesinfinity, i thought we already know that cos^2x= 1-sin^2x

Answer 14

well the way i did is much simpler an easier to get the answer
+ let me check the procedure @Michele_Laino did

Answer 15

more explanation:
\[(1+\sin x)^{2}-3(1+\sin x)(1-\sin x)=0\]
then I factor out (1+sin x):
\[(1+\sin x)[(1+\sin x-3(1-\sin x)]=0\]

Answer 16

hmm seems fine with me
you get 3 answers with that if my calculation are correct

Answer 17

well now you two have figured it out and thanks for trying to help but i am still lost. I have a lot of other problems for my midterm study guide to do so thanks anyway

Answer 18

after a simple simplification, we have:
\[4(1+\sin x)(\sin x-\frac{ 1 }{ 2 })=0\]
so please you have to solve this subsequent equations:
\[\sin x+1=0\]
and:
\[\sin x-\frac{ 1 }{ 2 }=0\]

Answer 19

in mine suggestion you have \(\cos(\pi/6)\cos x- \sin(\pi/6)\sin x=1/2==>\cos(\pi/6+x)=1/2 \)

Answer 20

not and! it is or :P

Answer 21

@0400782 did you understand what he did