Question

Find A, if 0°≤ A ≤ 90° and (i) 4sin² A - 3 = 0 (ii) 2cos² A + cos A - 1 = 0

Answer 1

( i ) :

$4 \: {sin}^{2} A \: - 3 = 0 \\ \\ \\ = > 4 \: {sin}^{2} A \: = 3 \\ \\ \\ = > {sin}^{2} A = \frac{3}{4} \\ \\ \\ = > sin \: A = \sqrt{ \frac{3}{4} } \\ \\ \\ = > sin \: A = \frac{ \sqrt{3} }{2} \\ \\ \\ = > sin \: A = sin \: 60 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: | \bold{ \frac{ \sqrt{3} }{2} = sin60} \\ \\ \\ = > A = 60 \degree$





( ii ) :





Let cosA = x
Then,


↪ 2 cos² A + cosA - 1 = 0

↪ 2 x² + x - 1 = 0

↪ 2 x² + 2x - x - 1 = 0

↪ 2x( x + 1 ) - ( x + 1 ) = 0

↪ ( x + 1 ) ( 2x - 1 ) = 0

↪ x = – 1 or x = ( 1 ÷ 2 )




If we take the negative value of x ( cosA ), anglw will be more than 90° which will be wrong as given in the question that the value of A is lesser than 90°


So, taking positive value that is ( 1 ÷ 2 )





$= > x = \frac{1}{2} \\ \\ = > cosA = cos60 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: | \bold{ \frac{1}{2} = cos60} \\ \\ = > A = 60 \degree$