trigonometry equation, solve exactly.

Question

Solve exactly between 0<x<2pi

2sin²x+sin(2x)+cos(2x)=0

Please and thank you!!!!

Answer 1

Given:  0<x<2pi .  This inequality means:  x>0 but x<2pi .

 

2Sin^2x+Sin(2x)+Cos(2x)=0  (equation given; when I write Sin^2x, it is the same as "Sin-squared x".

 

Now, Cos(2x) = 1-2Sin^2x (Trig. Formula)

 

So, 2Sin^2(x)+Sin(2x)+(1-2Sin^2(x))=0

 

      Or, Sin(2x)+1=0

 

      Or Sin(2x)=-1

 

      Or 2x=arcsin(-1)

   

      Or  x=1/2arcsin(-1)=1/2(3/2Pi)=3/4Pi

 

Answer:   x = (3/4) Pi 

 

 

P.S.:  Please get a 'Trigonometry Cheat Sheet' from Google Search and print it out. Cram all formulas!, like the one I have used.

 

 

           

 

      

      

 

    

      

           

 

       

 

Answer 2

2sin^2x+sin2x+cos2x=0

sin2x=2sinxcosx and cos2x=cos^2x-sin^2x are a trig identities

substitute

2sin^2x+2sinxcosx+cos^2x-sin^2x=0

combine similar terms

sin^2x+2sinxcosx+cos^2x=0

factor

(sinx+cosx)(sinx+cosx)=0

sinx+cosx=0

cosx=-sinx

divide both sides by cosx

1=-sinx/cosx

multiply both sides by -1

-1=sinx/cosx

-1=tanx

tanx=-1

tangent is negative in the 2nd and4th quadrant

tan45º=1 but we want -1

45º+90º=135º

tan135º=-1 (one answer)

45º+270º=315º (not "add 180" because we have to get to the 4th quadrant)

tan315º=-1 (second answer)

you can also look on the unit circle and see where sin and cos are opposites of each other because sin/cos=tan and a number divided by its opposite is -1

this would be 3∏/4 (135º) and 7∏/4 (315º)