Question

How do I solve for the value of x in degrees if √2sinx-1=0?

Answer 1

`x=135°+360°*n` or `x=360°*n+45°`

Explanation:

First, let's solve it in radians

`sqrt2sinx-1=0`
`therefore sinx=1/sqrt2=sqrt2/2`
If you look at Special Ratio, `sin45°` is `sqrt2/2`
proof)

SOH CAH TOA, `sin45°=(Opposite)/(Hypotenuse)`
`therefore 1/sqrt2=sqrt2/2`

But this was the only case when `0°<=x<=90°(0<=x<=pi/2)`

Now we need to look on real number range

Since `sin` has period of `2pi`, (`f(?)=f(?+2pi)`)

`thereforex=360°*n+45°(2npi+pi/4)`

But we know that `sin` function is positive @ 2nd quadrant.`(-, +)` Which means,

We need to find out the `sinx=sqrt2/2` for 2nd quadrant `(90°<=x<=180°(pi/2<=x<=pi))`
`x=(3pi)/4`
But `sinx` has period of `2pi`,

`therefore x=(3pi)/4+2npi`

If you write them in degrees,

`x=(3*180°)/4+360°*n=135°+360°*n`

`therefore x=135°+360°*n`