Question

Solve tan x < 2 sin x for [0,2pi)

Answer 1

tan x < 2 sin x
(sin x / cos x) < 2 sin x
sin x < 2 sin x cos x
sin x < sin 2x

What next?

Answer 2

i dont think you can multiply by cosx as it can be negative

Answer 3

cosine is not positive on \((0,2\pi)\) so you cannot multiply both sides by cosine

Answer 4

\[\frac{\sin x}{\cos x}-2\sin x <0\]
\[\sin x(\frac{1}{\cos x}-2)<0\]

Answer 5

\[tanx<2sinx\]

\[tanx - 2sinx < 0\]

\[\frac{(sinx - 2sinxcosx)}{cosx} < 0\]

now consider critical values where denominator, numerator = 0

we have to consider

sinx(1- 2cosx) = 0
cosx = 0

so get these values of x, then find the sign of \[\frac{(sinx - 2sinxcosx)}{cosx}\]

between them. this will solve your inequality.

Answer 6

have to make it positive or negative. what mertsj said

Answer 7

alternatively just solve tanx = 2sinx

then find the sign of tanx - 2sinx