The number of roots of equation x+2tanx=pi/2 in the interval[0,2pi]l
Answers
Three roots of Equation in the interval[0,2π] for x + 2Tanx = π/2
Step-by-step explanation:
x + 2Tanx = π/2
2Tanx = π/2 - x
Case 1 : when x lies between 0 to π/2
=> π/2 - x is + ve
2Tanx is also +ve between 0 to π/2
one function is increasing while other is decreasing
so one value will satisfy between 0 to π/2
Case 2 : when x lies between π/2 to π
=> π/2 - x is - ve
2Tanx is also -ve between π/2 to π
so one value will satisfy between 0 to π/2
Case 3 : when x lies between π to 3π/2
=> π/2 - x is - ve
But 2Tanx is +ve between π to 3π/2
so no such value
Case 4 : when x lies between 3π/2 to 2π
=> π/2 - x is - ve
2Tanx is also -ve between 3π/2 to 2π
so one value will satisfy between 3π/2 to 2π
so three such value exist for x in the interval[0,2π]
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