Find the number of the solutions of equation 99sin^2x-101=-sin^2x.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equation given is 99sin^2x-101=-sin^2x. The number of roots of the equation have to be determined.

99(sin x)^2 - 101 = -(sin x)^2

=> 100(sin x)^2 = 101

=> (sin x)^2 = 101/100

=> sin x = sqrt (101/ 100)

As sqrt (101/100) is greater than 1, and sin x can only take values in [ -1 , 1], there are no solutions.

99(sin x)^2 - 101 = -(sin x)^2 does not have any solutions

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First, we'll isolate (sin x)^2 to the left side. For this reason, we'll shift (sin x)^2 to the left side and we'll move -101 to the right, as 101:

99 (sin x)^2 + (sin x)^2 = 101

100 (sin x)^2 =101

We'll divide by 100:

(sin x)^2 = 101/100

sin x = sqrt (1.01)

But sqrt (1.01)>1 and -sqrt (1.01) < -1, which is impossible because the lower limit value of the sine function is -1 and the upper limit value is 1.

Therefore, the given equation has no solution.

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