Question #41b0c

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Answer 1 · 2017-11-18T00:08:55

$x = 60°, 300°$

Based on the pythagorean identity:

$cos^2x+sin^2x=1$

If we solve for $sin^2x$ , we get:

$sin^2x = 1 - cos^2x$

If we subsititute this value of $sin^2x$ into the original equation we get:

$4(1-cos^2x) = 1+4cosx$

If we distribute and move everything to the right side we get:

$0=4cos^2x + 4cosx-3$

Now we can factor this:

$0 = (2cosx-1)(2cosx+3)$

Set both sides equal to 0 and solve:

$2cosx-1 = 0$
$cosx = 1/2$
$x = +-arccos(1/2)$
$x = 60°, -60° = 60°, 300°$

$2cosx + 3=0$
$cosx = -3/2$
$x = +-arccos(-3/2)$
$NRS$

$therefore x = 60°, 300°$