Solve the equation $3sin^2x+cos^2x-1=0$ for $x in [0, 2pi]$ ?

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Answer 1 · 2018-05-18T16:36:49

${ 0, pi, 2 pi }$

$3sin²(x)+cos²(x)-1=0$
Because $sin²(x)+cos²(x)=1 <=> cos²(x)=1-sin²(x)$ ,

$3sin^2(x)+1-sin^2(x)-1=0$

$2sin^2(x)=0$

$sin(x)=0$

$x in {kpi, k in ZZ}$

Solve the equation 3sin^2x+cos^2x-1=03sin^2x+cos^2x-1=0 for x in [0, 2pi]x in [0, 2pi] ?