Solve on the interval [0,2pi]: 2sin(x)cos(x) + sqrt(2)cos(x) < 0?

Question

I have no idea how to solve this! I know 2sinxcosx is sin2x by definition, but...where should I go from there?

Answer 1

7 i think

Answer 2

Factor out cos(x) to get cos(x)*[2sin(x) + sqrt(2)] &lt; 0.

Now...cos(x) &lt; 0 on [pi/2, 3pi/2].

2sin(x) + sqrt(2) &lt; 0 on [4pi/3, 5pi/3].

So...the whole equation will be &lt; 0 when only one part or the other is &lt; 0.

f(x) &lt; 0 when x is in [pi/2, 4pi/3] u [5pi/3, 3pi/2].