Trigonometric equation involving (sin x ) ^ 2

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• Oct 8th 2012, 11:02 PM
nks2427
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
• Oct 8th 2012, 11:16 PM
MarkFL
Re: Trigonometric equation involving (sin x ) ^ 2
We are given to solve:

$\displaystyle \sin^2(x)=\sin(x)\cos(x)$ where $\displaystyle 0^{\circ}\le x\le360^{\circ}$

$\displaystyle \sin^2(x)-\sin(x)\cos(x)=0$

$\displaystyle \sin(x)(\sin(x)-\cos(x))=0$

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

$\displaystyle \sin(x)=0$

$\displaystyle x=0^{\circ},180^{\circ},360^{\circ}$

$\displaystyle \sin(x)-\cos(x)=0$

$\displaystyle \sin(x)=\cos(x)$

$\displaystyle x=45^{\circ},225^{\circ}$

The extra solutions you gave satisfy $\displaystyle \sin(x)+\cos(x)=0$, which is not part of the original equation.
• Oct 11th 2012, 10:21 AM
nks2427
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 ?
• Oct 11th 2012, 10:49 AM
skeeter
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
• Oct 11th 2012, 10:58 AM
nks2427
Re: Trigonometric equation involving (sin x ) ^ 2
Yeah ..should be more careful about these mistakes next time . Thank you dude : ))
• Oct 11th 2012, 11:08 AM
MaxJasper
Re: Trigonometric equation involving (sin x ) ^ 2
$\displaystyle x= \left\{0, \frac{\pi }{4}, \pi ,\frac{5\pi }{4}\right\}$