Solve for x in radians: sin 2x = cos x?

Question

I approached this problem by first using the double angle formula.
sin 2x = 2 sin x cos x = cos x

I then divided each side by cos x...
2 sin x = 1

...and then divided each side by 2.
sin x = 1/2

So x = pi/6 or 5*pi/6.

However, when I checked the answers, the solution set also contained pi/2 and 3*pi/2.

I realized then that instead of dividing each side by cos x, I should have subtracted and factored.
2 sin x cos x - cos x = 0
cos x (2 sin x - 1) = 0
cos x = 0 OR 2 sin x - 1 = 0
cos x = 0 OR sin x = 1/2
x = pi/2, 3*pi/2 OR x = pi/6, 5*pi/6

My questions are: Why was my first solution not complete? How can I make it complete? Is there some rule against dividing by cos x (or sin x, tan x, etc)? Or do I need to consider if x is negative?

Answer 1

What is cos(pi/2), or cos(3/2 pi)?


Hint: you can't divide by it.

Answer 2

> Why was my first solution not complete?

because it didn't include pi/2 or 3*pi/2.

> How can I make it complete?

the way you did; by factoring instead of cancelling cos x.

> Is there some rule against dividing by cos x (or sin x, tan x, etc)?

there's the general rule that you can never divide by zero. if you ever divide by some function of x, you must ensure that the function cannot take on a value of zero before dividing.

> Or do I need to consider if x is negative?

aha. your question should say to find all solutions for x in the interval [0, 2*pi]. if not, your book's answer is wrong; there are infinitely many solutions, including infinitely many solutions of negative values of x.

in particular, since sin(x + 2*pi) = sin x and cos(x + 2*pi) = cos x, if you have any particular solution x, then x + 2*pi is also a solution (and x - 2*pi). there may even be more (sorry, i haven't fully analyzed your problem).

it's a good exercise to find *all* solutions, and write the solution set in a form mathematicians use.

HTH & HAND!

Answer 3

To answer your question, you needed to consider the divide by zero situations. That would've given you the other two solutions. cos(pi/2) = cos(3 pi/2) = 0, just as sin(pi) and sin(3 pi) = 0.

However, I think I would've approached the problem differently. I'd have drawn a right triangle, labeled one angle x, and the other angle 2x. Right away I'd see a 30-60-90 triangle, and by looking at sine and cosine relationships the other quadrants, I'd probably have gotten the other solutions as well.

Answer 4

sin2x = cosx;
Here it comes:
2 sinx cosx = cosx

answer: cosx = 0 => x = pi/2 + pi*k; where k=0, +-1, +-2, ... etc.
and
sinx = 1/2 => x = pi/6 + 2*pi*k
or x = 5*pi/6 + 2*pi*k; where k is the same...

It's easy as 1 2 3.... :)

Answer 5

Hey If you divide by cos x you are loosening solutions. I will explain this with this example, that is not trigonometric, but will do the job:

0x4 = 7x0, because 0 = 0. But I cant cancel these 0, because I will get 4 = 7.

Ana

Answer 6

because u cannot cancel the variable from the equation
it's like (x-1)^3=0
u can't sayy the equation has 1 root
it has 3 roots and all three are 1
what u did is that u removed the variable from the equation

Answer 7

sin2x=cosx
2sinxcosx=cosx
sinx=1/2 and sinx=sinpi/6
so the general solution is n*pi+(-1)n*pi/6 where nEZ