Tough Trig Identities Problem Help?

Let's look first at a simpler problem.

sin(θ) + cos(θ) = 0

This only happens when θ is in the second or fourth quadrant, because cos and sin are opposite signs in those quadrants. Looking at the second quadrant, this can only happen at 3π/4. We know this because that is where

-sin(θ) = cos(θ)

We have this valid at one point, but we also know that sin and cos are periodic with a period of 2π. Therefore:

θ = 3π/4 + 2nπ

where n is any integer (positive or negative).

Now let's look at the fourth quadrant. The only place in the fourth quadrant where

-sin(θ) = cos(θ)

is true is at 7π/4. Using the 2π period, we end up with:

θ = 7π/4 + 2nπ

for any integer n. If we look closely, we see that

θ = 3π/4 + 2nπ

θ = 7π/4 + 2nπ

Only differ by π. We can simplify the two expressions into:

θ = 3π/4 + nπ

for any integer n.

Your problem is

sin(3x) + cos(3x) = 0

If we set θ = 3x we end up with the last problem where we found:

θ = 3π/4 + nπ

Using θ = 3x, we get:

3x = 3π/4 + nπ

Solving for x, we get:

x = (3π/4 + nπ)/3

x = π/4 + nπ/3

Since your interval is from 0 to 2π, we need to find our range for n. This is done by setting x equal to the end points. For x = 0:

x = 0 = π/4 + nπ/3

nπ/3 = -π/4

n = -3/4

Since this was for x = 0 and we cannot go below x = 0, we round to the next valid n, which is n = 0. For x = 2π

x = 2π = π/4 + nπ/3

nπ/3 = 2π - π/4

nπ/3 = 7π/4

n = 3*7/4

n = 21/4

n = 5.25

Since this is for our upper limit, we round down to n = 5.

This tells us for n=0 to 5 that our answer is

x = π/4 + nπ/3

or

x = π/4 + 0*π/3, π/4 + 1*π/3, π/4 + 2*π/3, π/4 + 3*π/3, π/4 + 4*π/3, π/4 + 5*π/3

x = π/4, π/4 + π/3, π/4 + 2π/3, π/4 + π, π/4 + 4π/3, π/4 + 5π/3

x = π/4, 7π/12, 11π/12, 5π/4, 19π/12, 23π/12