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Solving a trig equation

I have to solve:

2sinx = cos(x-60), for 0<=x<=360

so far i expanded the cos part

2sinx = cosxcos60 + sinxsin60

as cos60 = 1/2 and sin60 = (sqrt3)/2

2sinx = (1/2)cosx + ((sqrt3)/2)sinx

sinx = cosx + sqrt3 sinx

(1-sqrt3)sinx = cosx

but here's where I'm stuck, i have 2 trig functions in one equation and therefore cannot solve.
 

cristo

Staff Emeritus
Science Advisor
8,056
72
I have to solve:

2sinx = cos(x-60), for 0<=x<=360

so far i expanded the cos part

2sinx = cosxcos60 + sinxsin60

as cos60 = 1/2 and sin60 = (sqrt3)/2

2sinx = (1/2)cosx + ((sqrt3)/2)sinx

sinx = cosx + sqrt3 sinx
This line is incorrect. You have multiplied both sides by 2 to remove the factor of 1/2 on the right hand side, so the left hand side should be premultipled by 4.
 
ahh yes, how silly of me,

but even now i have:

(4 - sqrt3)sinx = cosx

i need it all in sins or cos'
 
but if i divide through by cosx, i'll loose that solution of x because the cos' will cancel, its an equation not an expression.
 

cristo

Staff Emeritus
Science Advisor
8,056
72
No you won't. There's only going to be one solution to that equation in the range specified.
 
the answer in the back of the text book gave 2 solutions.
I was told to never divide through with an equation, only with an expression

perhaps you only got one solution because you divided through to solve.
 
Last edited:

cristo

Staff Emeritus
Science Advisor
8,056
72
perhaps you only got one solution because you divided through to solve.
Sorry, Im being really stupid! You don't lose a solution by dividing by cos(x), but of course the function tan(x)= 1/(4-sqrt(3)), is periodic with period 180 degrees. So, the solutions to this in the given range will be the principal value for arctan(1/(4-sqrt(3))) [the one given by your calculator], and this value with 180 added on.
 
ok so arctan 1/(4-sqrt3) gives me my priciple 23.8, and then +180 to give 203.8, my secondary.

Solved!
cheers, im a bit weary about dividing through but im not gonna argue with the outcome!
 
206
0
It's fine because you know that an x where cos(x) is zero can't possibly be a solution (otherwise you have 4 - sqrt(3) = 0).
 
286
0
If you really wanted to, when you had sin on one side and cos on the other side, you could square both sides (and possibly introduce extraneous roots), and use an identity for (sinx)^2 or (cosx)^2, changing it to a quadratic type. The method above is much easier.
 

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