Question

How do you solve `10cos(x)^2-3sin(x)+9=0`?

Answer 1

`sinx= 0.2`
`sinx= -0.5`

Explanation:

Well according to me the equation given by you has a slight mistake the real equation is `-10cos^2x - 3sinx + 9 = 0`, notice the negative sign in front of 10.

Now this equation can be solved by converting `cos^2x` into `(1-sin^2x)` and proceed as follows:

`-10(1-sin^2x)+3sinx+9=0`
`-10-10sin^2x+3sinx+9=0`
`-10sin^2x+3sinx-1=0`
`10sin^2x-3sinx-1=0`

Now you can solve for `sinx` by middle term splitting—you would get two values:

1. `sinx = 1/2`

2. `sinx=-1/5`