Question

2sino = x + 1/x , then x = ?​

Answer 1

Theta is written as A.

Answer:

sinA ± cosAi

Step-by-step explanation:

Here,

⇒ 2sinA = x + 1 / x

⇒ 2sinA = ( x^2 + 1 ) / x

⇒ 2x.sinA = x^2 + 1

⇒ x^2 + 1 - 2x.sinA = 0

⇒ x^2 - 2x.sinA + 1 = 0

Using Quadratic Formula :

   for an equation ax^2 + bx + c = 0, x = [ - b ± √( b^2 - 4ac ) ] / 2a

Thus, here,

⇒ x = [ - ( - 2sinA ) ± √{ ( - 2sinA )^2 - 4( 1 )( 1 ) } ] / 2

⇒  x = [ 2sinA ± √( 4sin^2 A - 4 ) ] / 2

⇒ x = [ 2sinA ± √{ 4( sin^2 A - 1 ) } / 2

⇒ x = [ 2sinA ± √{ - 4cos^2 A } ] / 2      { 1 - sin^2 A = cos^2 A }

⇒ x = [ 2sinA ± 2cosA√( - 1 ) ] / 2

⇒  x = sinA ± cosAi     { where i represents iota }

Answer 2

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