How to Solve $5-4cosx=4sin^2x$ for $-90°<=x<=90°$ ?

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$5-4cosx=4sin^2x$ for $-90°<=x<=90°$

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Answer 1 · 2017-05-25T01:40:51

$x=60^@,-60^@$

Identity Required:
$sin^2 x+cos^2 x=1$
$sin^2 x=1-cos^2 x$

Now solving:
$5-4cosx=4sin^2 x$
$5-4cosx=4(1-cos^2 x)$
$5-4cosx=4-4cos^2 x$
$4cos^2 x -4cos x+1=0$
$(2cosx -1)^2=0$
$cosx=1/2$
$x=60^@,-60^@$