Question

I need ans anyone pls help me​

Answer 1

2sin² A = 4 + 3cos A

=> sin² A + cos² A = 1

=> sin² A = ( 1 - cos² A )

=> 2 ( 1 - cos² A ) = 4 + 3 cos A

=> 2 - 2cos² A = 4 + 3 cos A

=> 2 cos² A + 3 cos A + 2 = 0

Discriminant of the above equation is ,

=> √ { (3)² - 4 (2)(2) }

=> √ { 9 -16 }

=> √ -7 = √7 i

so , the above equation has NO REAL ROOTS .

Therefore , there are 0 values of A in the interval [ 0 , 2π ] .

Answer 2

Answer:

the answer if this question is

1