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











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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




















  • 1




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












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










  • Whoops, my bad.
    – Shubham Johri
    Nov 27 at 8:33















up vote
1
down vote

favorite













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




















  • 1




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












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










  • Whoops, my bad.
    – Shubham Johri
    Nov 27 at 8:33













up vote
1
down vote

favorite









up vote
1
down vote

favorite












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
















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 at 8:01









TheSimpliFire

11.9k62257




11.9k62257










asked Nov 27 at 7:50









D Tiwari

5,3132630




5,3132630








  • 1




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












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










  • Whoops, my bad.
    – Shubham Johri
    Nov 27 at 8:33














  • 1




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












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










  • Whoops, my bad.
    – Shubham Johri
    Nov 27 at 8:33








1




1




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






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














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




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












Whoops, my bad.
– Shubham Johri
Nov 27 at 8:33




Whoops, my bad.
– Shubham Johri
Nov 27 at 8:33










2 Answers
2






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2
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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






























    up vote
    1
    down vote













    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























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











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    2 Answers
    2






    active

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    2 Answers
    2






    active

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    active

    oldest

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    active

    oldest

    votes








    up vote
    2
    down vote













    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



























      up vote
      2
      down vote













      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

























        up vote
        2
        down vote










        up vote
        2
        down vote









        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














        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 at 8:02

























        answered Nov 27 at 7:55









        gimusi

        93k94495




        93k94495






















            up vote
            1
            down vote













            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























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















            up vote
            1
            down vote













            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























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













            up vote
            1
            down vote










            up vote
            1
            down vote









            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














            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 at 8:09

























            answered Nov 27 at 8:00









            TheSimpliFire

            11.9k62257




            11.9k62257












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


















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
















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




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


















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