Solve the equation $2cos2x+1/2sin2x=sin^2x$ ?

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Answer 1 · 2017-10-12T05:32:37

$x=npi+pi/4$ or $x=npi+tan^(-1)(-2/3)$ , where $n$ is an integer

$2cos2x+1/2sin2x=sin^2x$ can be written as

$2(1-2sin^2x)+sinxcosx=sin^2x$

or $5sin^2x-sinxcosx-2=0$ and dividing each term by $cos^2x$

$5tan^2x-tanx-2(1+tan^2x)=0$

or $3tan^2x-tanx-2=0$

i.e. $(3tanx+2)(tanx-1)=0$

i.e. either $tanx-1=0=>tanx=1=tan(pi/4)$ i.e. $x=npi+pi/4$

or $3tanx+2=0=>tanx=-2/3$ i.e. $x=npi+tan^(-1)(-2/3)$ , where $n$ is an integer

Solve the equation 2cos2x+1/2sin2x=sin^2x2cos2x+1/2sin2x=sin^2x?