# How to solve 2sin x > √ 5 cos x ?

jeew-m | Certified Educator

2sinx>√5cosx

2sinx-√5cosx >0

Assume a triangle ABC where angle B is 90 deg.

Let AB = 2 and BC =√5

So AC = (√(2^2+(√5)^2)) =3  (AC>0)

Then ;

sinA = √5/(√(2^2+(√5)^2)

= √5/√ (4+5)

= √5/3

CosA = 2/(√(2^2+(√5)^2))

= 2/3

So

2sinx-√5cosx

= (√(2^2+(√5)^2)){[2/(√(2^2+(√5)^2)]*sinx-[√5/(√(2^2+(√5)^2))]*cosx

= 3(cosA.sinx-sinA.cosx)

We know  sin(A-B)=sin A cos B - cos A sin B

So 2sinx-√5cosx = 3(cosA.sinx-sinA.cosx)

= 3 (sin(x-A)

Sin A = √5/3

A = sin-1(√5/3)

2sinx-√5cosx = 3 (sin(x-0.841)

So our equation become;

3 (sin(x-0.841)>0

(sin(x-0.841)>0

x-0.841 >0+n*pi

x > n*pi+0.841 where n is any integer

sciencesolve | Certified Educator

Notice that the division by cos x is possible since `cosx != 0` , hence, you should divide both sides by `cos x`  such that:

`2 sin x / cos x gt sqrt5 cos x/ cos x`

`2 tan x gt sqrt 5 =gt tan x gt sqrt5/2`

`x gt tan^(-1) (sqrt5/2) + n pi`

`xgt 48^o + 180^o*n`

Hence, evaluating the general solution to trigonometric inequality yields `xgt 48^o + 180^o*n` .