How to solve 2sin x > √ 5 cos x ?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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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` .

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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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)

        = 0.841 rad

 

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

    

Sources:

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