Question #9c5a0

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Answer 1 · 2016-05-05T09:02:25

$x = pi/4+pik$ , $k in ZZ$

$1+sin^2(x) = 3sin(x)cos(x)$ , $tan(x)!=1/2$

$=> (sin^2(x)+cos^2(x))+sin^2(x) = 3sin(x)cos(x)$

$=> cos^2(x) + 2sin^2(x) = 3sin(x)cos(x)$

$=> cos^2(x)-3sin(x)cos(x)+2sin^2(x) = 0$

$=> (cos(x)-sin(x))(cos(x)-2sin(x)) = 0$

$=> cos(x)-sin(x) = 0$ or $cos(x) - 2sin(x) = 0$

As the second equation may be expressed as $sin(x)/cos(x)=1/2$ , violating our initial condition of $tan(x)!=1/2$ , we will only focus on the first.

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

$=> cos(x) = sin(x)$

$:. x = pi/4+pik$ , $k in ZZ$