Question

If 3(1+sinx) _> 1 + cos2x,
X belongs to [0 to pi]
Then the number of value of x is?

Answer 1

Answer:

all values of x between 0 to π

Step-by-step explanation:

If 3(1+sinx) _> 1 + cos2x,

X belongs to [0 to pi]

Then the number of value of x is?

3(1+sinx) ≥ 1 + cos2x,

=> 3(1+sinx) ≥ 1 + 1 - 2Sin²x

=> 1 + 3Sinx ≥ - 2Sin²x

=> 2Sin²x + 3Sinx + 1 ≥ 0

=> (2Sinx + 1)(Sinx + 1) ≥ 0

=> Always Sinx + 1  ≥ 0

=> 2Sinx + 1 ≥ 0

=> Sinx ≥ -1/2

Which is true for all values of x between 0 to π

Answer 2

look bro I've done what I could do. Plij understand.