# Solve for y 11 sin^2y=13-sin^2y where 0 < y < 360. How many angles are the solution of the equation.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Solve for y  11 sin^2y=13-sin^2y where 0 < y < 360.

Add sin^2y to both sides of the equation:

11sin^2y +sin^2y = 13-sin^2y+sin^2y.

12sin^2y = 13.

sin^2y = 13/12.

But sin^2y  > 1 is not admissible as the siney functin is defined only for -1 <= siny <= 1, sin^2 y > 1 is not possible. There is no solution for y.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The first step is to isolate (sin y)^2 to the left side. For this reason, we'll add (sin y)^2 both sides:

11 (sin y)^2 + (sin y)^2 = 13

12 (sin y)^2 =13

We'll divide by 12:

(sin y)^2 = 13/12

sin y = sqrt (13/12)

But sqrt (13/12)>1 and -sqrt (13/12) < -1, which is impossible because the limit values of the sine function are -1 and 1.

So, there are no angles to satisfy the given equation, or, more precise, the equation has no solution.