Find all values of x in the interval [0,2 pi] that satisfy sin x = tan x.?

Question

This problem seems so easy but I can't seem to figure it out. I can see what the answer is from looking at a graph of the two functions but when I try to solve it without the graphs I get that x=0,2 pi.

I started by recognizing that tan x = sin x/cos x and then multiplied both sides by cos x/sin x (the reciprocal), which gave me cos x = 1 and cos x =1 at 0 and 2 pi. So I don't understand, without the graph, why the answer is x = 0, pi, 2 pi.

Answer 1

Your first step is good:

sin x = sin x / cos x

Your second step, "Multiply by cos x / sin x", is not quite good. If it happens that sin x = 0, then you have just multiplied by something divided by zero, which is not allowed. You can solve the problem after multiplying by cos x / sin x, but you need to make a separate case considering what happens when sin x = 0. (Because in that case, multiplying by cos x / sin x is not allowed).

Here is how I might right up a solution.

sin x = sin x / cos x

CASE I: sin x is not equal to 0. Then we are allowed to multiply by (cos x) / (sin x), which gives us

cos x = 1

This gives the solutions x = 0, 2pi.

CASE II: sin x = 0. Then

0 = 0/ cos x

This holds provided cos x is not zero, but it never is when sin x = 0. So any time sin x = 0, x is also a solution. So we get the additional solution

x = pi.

(Of course, sin x = 0 at 0 and 2pi also, but we already got those solutions from case I.)

Answer 2

cos x is also 1 at x = pi, I don't see what your problem is

sin x = 0 at x = pi
tan x = 0 at x =pi since it is sinx/cosx = sinx/SQRT(1 - sin^2x) and so it will also be 0 whenever sinx is 0