Trig Question

  • #1
frenzal_dude
77
0

Homework Statement


Hi, I need to solve for x:
[tex]\sqrt{3}cos(x)+sin(x)=1[/tex]


Homework Equations





The Attempt at a Solution


[tex]3(Cosx)^{2}+2\sqrt{3}CosxSinx+(Sinx)^{2}-1=0[/tex]
[tex]3(Cosx)^{2}+2\sqrt{3}CosxSinx+(Cosx)^{2}=[/tex]
[tex]4(Cosx)^{2}+2\sqrt{3}SinxCosx=0[/tex]
[tex]Cosx(4Cosx+2\sqrt{3}Sinx)=0[/tex]
[tex]\therefore Cosx=0[/tex] x=90 or 270.
OR
[tex]4Cosx=-2\sqrt{3}Sinx[/tex]
I wasn't sure how to work out that last bit.
Hope you guys can help.
Frenzal
 

Answers and Replies

  • #2
36,338
8,295

Homework Statement


Hi, I need to solve for x:
[tex]\sqrt{3}cos(x)+sin(x)=1[/tex]


Homework Equations





The Attempt at a Solution


[tex]3(Cosx)^{2}+2\sqrt{3}CosxSinx+(Sinx)^{2}-1=0[/tex]
[tex]3(Cosx)^{2}+2\sqrt{3}CosxSinx+(Cosx)^{2}=[/tex]
In the line above you replaced sin^2(x) - 1 with cos^2(x). That should be -cos^2(x).

Also, you lost the 0 on the right-hand side.
[tex]4(Cosx)^{2}+2\sqrt{3}SinxCosx=0[/tex]
[tex]Cosx(4Cosx+2\sqrt{3}Sinx)=0[/tex]
[tex]\therefore Cosx=0[/tex] x=90 or 270.
OR
[tex]4Cosx=-2\sqrt{3}Sinx[/tex]
I wasn't sure how to work out that last bit.
Hope you guys can help.
Frenzal
Use the identity that sin(x)/cos(x) = tan(x). You will first need to fix the error noted above, though.

Also, be sure to check your solutions in the original equation. By squaring both sides, you might be introducing extraneous solutions: solutions of your squared equation that are not solutions of the original equation.

One other thing. Since there are no restrictions on x, there are going to be an infinite number of solutions. For example, if x = pi/6 were to turn out to be a solution, then pi/6 + n*2pi, n = 0, +/-1, +/-2, ... would represent all such solutions.
 
  • #3
graphene
195
2
divide both sides of the eqn by 2.
& then u can write that as
sin60*cosx + cos60*sinx = 1/2
sin(x + 60) = sin(30) = sin(150)
x = -30, 90

& then substitute & check which is correct
 

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