Trigonometric equation involving (sin x ) ^ 2

So this is the equation,

(sin x)^2 = sinx cosx for x lying between 0 and 360 degrees both inclusive.

Solving it, i got the following values for x

x = 0, 45 , 135 , 180 , 225 , 315 , 360

but the correct values for x are x = 0, 45 , 180 , 225 , 360

I dont understand why 135, 315 is not considered as correct .

Nid ur help guys. thnks in advance

Re: Trigonometric equation involving (sin x ) ^ 2

We are given to solve:

where

Now, in turn equate both factors to zero, and solve for on the given interval:

The extra solutions you gave satisfy , which is not part of the original equation.

Re: Trigonometric equation involving (sin x ) ^ 2

sin x = cos x

i solved it this way..

squaring both sides,

(sin x)² = (cos x)²

(sin x)² = 1 - (sin x)²

2 (sin x)² = 1

sin x = ± (1/√2)

i get x= 45,135 and x=225,315

how do i reject 135 and 315 ?

Re: Trigonometric equation involving (sin x ) ^ 2

Quote:

how do i reject 135 and 315 ?

you're expected to know that sine and cosine have opposite signs at x = 135 and x = 315

Re: Trigonometric equation involving (sin x ) ^ 2

Yeah ..should be more careful about these mistakes next time . Thank you dude : ))

Re: Trigonometric equation involving (sin x ) ^ 2