Real solution of $(cos x -sin x)cdot bigg(2tan x+frac{1}{cos x}bigg)+2=0.$












1












$begingroup$



Real solution of equation $$(cos x -sin x)cdot bigg(2tan x+frac{1}{cos x}bigg)+2=0.$$




Try: Using Half angle formula



$displaystyle cos x=frac{1-tan^2x/2}{1+tan^2 x/2}$ and $displaystyle sin x=frac{2tan^2 x/2}{1+tan^2 x/2}$



Substuting These values in equation



we have an polynomial equation in terms of $t=tan x/2$



So our equation $$3t^{4}+6t^{3}+8t^{2}-2t-3=0$$
Could Some Help me how to Factorise it.



OR is there is any easiest way How to solve it, Thanks






trigonometry polynomials roots







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $displaystyle sin x=frac{2tan x/2}{1+tan^2 x/2}$. Also, $x=pi$ is a real solution of the equation, but you can't get that from your substitution.
    $endgroup$
    – Shubham Johri
    Nov 27 '18 at 7:59












  • $begingroup$
    Are you sure $x=pi$ is a solution?
    $endgroup$
    – gimusi
    Nov 27 '18 at 8:10










  • $begingroup$
    Whoops, my bad.
    $endgroup$
    – Shubham Johri
    Nov 27 '18 at 8:33
















1












$begingroup$



Real solution of equation $$(cos x -sin x)cdot bigg(2tan x+frac{1}{cos x}bigg)+2=0.$$




Try: Using Half angle formula



$displaystyle cos x=frac{1-tan^2x/2}{1+tan^2 x/2}$ and $displaystyle sin x=frac{2tan^2 x/2}{1+tan^2 x/2}$



Substuting These values in equation



we have an polynomial equation in terms of $t=tan x/2$



So our equation $$3t^{4}+6t^{3}+8t^{2}-2t-3=0$$
Could Some Help me how to Factorise it.



OR is there is any easiest way How to solve it, Thanks






trigonometry polynomials roots







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $displaystyle sin x=frac{2tan x/2}{1+tan^2 x/2}$. Also, $x=pi$ is a real solution of the equation, but you can't get that from your substitution.
    $endgroup$
    – Shubham Johri
    Nov 27 '18 at 7:59












  • $begingroup$
    Are you sure $x=pi$ is a solution?
    $endgroup$
    – gimusi
    Nov 27 '18 at 8:10










  • $begingroup$
    Whoops, my bad.
    $endgroup$
    – Shubham Johri
    Nov 27 '18 at 8:33














1












1








1





$begingroup$



Real solution of equation $$(cos x -sin x)cdot bigg(2tan x+frac{1}{cos x}bigg)+2=0.$$




Try: Using Half angle formula



$displaystyle cos x=frac{1-tan^2x/2}{1+tan^2 x/2}$ and $displaystyle sin x=frac{2tan^2 x/2}{1+tan^2 x/2}$



Substuting These values in equation



we have an polynomial equation in terms of $t=tan x/2$



So our equation $$3t^{4}+6t^{3}+8t^{2}-2t-3=0$$
Could Some Help me how to Factorise it.



OR is there is any easiest way How to solve it, Thanks






trigonometry polynomials roots







share|cite|improve this question











$endgroup$





Real solution of equation $$(cos x -sin x)cdot bigg(2tan x+frac{1}{cos x}bigg)+2=0.$$




Try: Using Half angle formula



$displaystyle cos x=frac{1-tan^2x/2}{1+tan^2 x/2}$ and $displaystyle sin x=frac{2tan^2 x/2}{1+tan^2 x/2}$



Substuting These values in equation



we have an polynomial equation in terms of $t=tan x/2$



So our equation $$3t^{4}+6t^{3}+8t^{2}-2t-3=0$$
Could Some Help me how to Factorise it.



OR is there is any easiest way How to solve it, Thanks






trigonometry polynomials roots




trigonometry polynomials roots






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 8:01









TheSimpliFire

12.4k62460




12.4k62460










asked Nov 27 '18 at 7:50









DXTDXT

5,6892630




5,6892630








  • 1




    $begingroup$
    $displaystyle sin x=frac{2tan x/2}{1+tan^2 x/2}$. Also, $x=pi$ is a real solution of the equation, but you can't get that from your substitution.
    $endgroup$
    – Shubham Johri
    Nov 27 '18 at 7:59












  • $begingroup$
    Are you sure $x=pi$ is a solution?
    $endgroup$
    – gimusi
    Nov 27 '18 at 8:10










  • $begingroup$
    Whoops, my bad.
    $endgroup$
    – Shubham Johri
    Nov 27 '18 at 8:33














  • 1




    $begingroup$
    $displaystyle sin x=frac{2tan x/2}{1+tan^2 x/2}$. Also, $x=pi$ is a real solution of the equation, but you can't get that from your substitution.
    $endgroup$
    – Shubham Johri
    Nov 27 '18 at 7:59












  • $begingroup$
    Are you sure $x=pi$ is a solution?
    $endgroup$
    – gimusi
    Nov 27 '18 at 8:10










  • $begingroup$
    Whoops, my bad.
    $endgroup$
    – Shubham Johri
    Nov 27 '18 at 8:33








1




1




$begingroup$
$displaystyle sin x=frac{2tan x/2}{1+tan^2 x/2}$. Also, $x=pi$ is a real solution of the equation, but you can't get that from your substitution.
$endgroup$
– Shubham Johri
Nov 27 '18 at 7:59






$begingroup$
$displaystyle sin x=frac{2tan x/2}{1+tan^2 x/2}$. Also, $x=pi$ is a real solution of the equation, but you can't get that from your substitution.
$endgroup$
– Shubham Johri
Nov 27 '18 at 7:59














$begingroup$
Are you sure $x=pi$ is a solution?
$endgroup$
– gimusi
Nov 27 '18 at 8:10




$begingroup$
Are you sure $x=pi$ is a solution?
$endgroup$
– gimusi
Nov 27 '18 at 8:10












$begingroup$
Whoops, my bad.
$endgroup$
– Shubham Johri
Nov 27 '18 at 8:33




$begingroup$
Whoops, my bad.
$endgroup$
– Shubham Johri
Nov 27 '18 at 8:33










2 Answers
2






active

oldest

votes


















2












$begingroup$

That’s a nice way to solve but recall that $sin x=frac{2t}{1+t^2}$.



I didn’t check whether it is only a typo or a mistake in the derivation but note that wolfy suggests $$p(t)=(3t^2-1)(t^2+2t+3)$$.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    HINT:



    Since $sin x=frac{2t}{1+t^2}$, the equation becomes $$left(frac{1-t^2-2t}{1+t^2}right)left(frac{2t+1+t^2}{1-t^2}right)+2=0implies -t^4-4t^3-4t^2+1=2t^4-2$$ or $$3t^4+4t^3+4t^2-3=0$$ from which a root is $t=-1$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      It is very far from a hint in that way and I think the asker can get it by his/herself.
      $endgroup$
      – gimusi
      Nov 27 '18 at 8:13











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    That’s a nice way to solve but recall that $sin x=frac{2t}{1+t^2}$.



    I didn’t check whether it is only a typo or a mistake in the derivation but note that wolfy suggests $$p(t)=(3t^2-1)(t^2+2t+3)$$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      That’s a nice way to solve but recall that $sin x=frac{2t}{1+t^2}$.



      I didn’t check whether it is only a typo or a mistake in the derivation but note that wolfy suggests $$p(t)=(3t^2-1)(t^2+2t+3)$$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        That’s a nice way to solve but recall that $sin x=frac{2t}{1+t^2}$.



        I didn’t check whether it is only a typo or a mistake in the derivation but note that wolfy suggests $$p(t)=(3t^2-1)(t^2+2t+3)$$.






        share|cite|improve this answer











        $endgroup$



        That’s a nice way to solve but recall that $sin x=frac{2t}{1+t^2}$.



        I didn’t check whether it is only a typo or a mistake in the derivation but note that wolfy suggests $$p(t)=(3t^2-1)(t^2+2t+3)$$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 27 '18 at 8:02

























        answered Nov 27 '18 at 7:55









        gimusigimusi

        92.8k84494




        92.8k84494























            1












            $begingroup$

            HINT:



            Since $sin x=frac{2t}{1+t^2}$, the equation becomes $$left(frac{1-t^2-2t}{1+t^2}right)left(frac{2t+1+t^2}{1-t^2}right)+2=0implies -t^4-4t^3-4t^2+1=2t^4-2$$ or $$3t^4+4t^3+4t^2-3=0$$ from which a root is $t=-1$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It is very far from a hint in that way and I think the asker can get it by his/herself.
              $endgroup$
              – gimusi
              Nov 27 '18 at 8:13
















            1












            $begingroup$

            HINT:



            Since $sin x=frac{2t}{1+t^2}$, the equation becomes $$left(frac{1-t^2-2t}{1+t^2}right)left(frac{2t+1+t^2}{1-t^2}right)+2=0implies -t^4-4t^3-4t^2+1=2t^4-2$$ or $$3t^4+4t^3+4t^2-3=0$$ from which a root is $t=-1$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              It is very far from a hint in that way and I think the asker can get it by his/herself.
              $endgroup$
              – gimusi
              Nov 27 '18 at 8:13














            1












            1








            1





            $begingroup$

            HINT:



            Since $sin x=frac{2t}{1+t^2}$, the equation becomes $$left(frac{1-t^2-2t}{1+t^2}right)left(frac{2t+1+t^2}{1-t^2}right)+2=0implies -t^4-4t^3-4t^2+1=2t^4-2$$ or $$3t^4+4t^3+4t^2-3=0$$ from which a root is $t=-1$.






            share|cite|improve this answer











            $endgroup$



            HINT:



            Since $sin x=frac{2t}{1+t^2}$, the equation becomes $$left(frac{1-t^2-2t}{1+t^2}right)left(frac{2t+1+t^2}{1-t^2}right)+2=0implies -t^4-4t^3-4t^2+1=2t^4-2$$ or $$3t^4+4t^3+4t^2-3=0$$ from which a root is $t=-1$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 27 '18 at 8:09

























            answered Nov 27 '18 at 8:00









            TheSimpliFireTheSimpliFire

            12.4k62460




            12.4k62460












            • $begingroup$
              It is very far from a hint in that way and I think the asker can get it by his/herself.
              $endgroup$
              – gimusi
              Nov 27 '18 at 8:13


















            • $begingroup$
              It is very far from a hint in that way and I think the asker can get it by his/herself.
              $endgroup$
              – gimusi
              Nov 27 '18 at 8:13
















            $begingroup$
            It is very far from a hint in that way and I think the asker can get it by his/herself.
            $endgroup$
            – gimusi
            Nov 27 '18 at 8:13




            $begingroup$
            It is very far from a hint in that way and I think the asker can get it by his/herself.
            $endgroup$
            – gimusi
            Nov 27 '18 at 8:13


















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