Solving Trigonomic Equations. How do I go about solving 4sin^4x + 3sin^2x -1 =0?

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Answer 1 · 2018-01-25T12:15:41

$4sin^4x + 3sin^2x -1 =0$

$=>4sin^4x + 4sin^2x-sin^2x -1 =0$

$=>4sin^2x(sin^2x+1) -(sin^2x +1) =0$
$=>(4sin^2x-1)(sin^2x+1) =0$

$(sin^2x+1) =0$ not possible

When

$(4sin^2x-1)=0$

$=>2(1-cos2x)=1$

$=>2-2cos2x)=1$

$=>2cos2x=1$

$=>cos2x=1/2=cos(pi/3)$

$=>2x=2npipmpi/3" where "n in ZZ$

$=>x=npipmpi/6" where "n in ZZ$