Question
has
- no solution
- one solution
- two solutions
- three solutions
Hint:
In this question, we have to find how many solution this equation does. The equation is . Check for , the value of x in all those quadrant where tan is positive and tan is negative.
The correct answer is: one solution
has
Here we have to find the how many of solution this equation does.
Firstly, we have equation,
We know x ∈ 2π
Lets spilt into two parts,
x ∈ [ 0 , π/2 ] U [π , 3 π/2 ] and [π/2 , π] U [3 π/2 , 2 π ]
At 1st case ,
x ∈ [ 0 , π/2 ] U [π , 3 π/2 ]
at this region tan is always positive,
then we can write,
|tanx| = tanx
So,
= 0
So it has no solution because, cosx in denominator and we know cosx ≠ 0 ..
At 2nd case,
x ∈ [π/2 , π] U [3 π/2 , 2 π ] ,
at this range tan is always negative,
so | tanx| = -tanx
we have,
-
-2 tanx =
= -1/2
sinx = -1/2
at [π/2 , π] sine is positive but it not possible ,
but at [3 π/2 , 2 π ] sine is negative,
so x = 330°
Hence it has only one solution,
Therefore , the correct answer is one solution.
In this question, we have to find the number of solution. In each quadrant tan value is different. Make two different case, one is where tan is positive ,[ 0 , π/2 ] U [π , 3 π/2 ] . And second case , tan is negative at [π/2 , π] U [3 π/2 , 2 π ] .
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