How do you solve $cos x + sin x tan x = 2$ $cosx+sinxtanx=2$ ?

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Answer 1 · 2015-05-25T18:43:49

$tan (x) = \frac{sin (x)}{cos (x)}$ $tan(x) = sin(x)/cos(x)$

So
$cos (x) + sin (x) tan (x) = 2$ $cos(x) + sin(x)tan(x) = 2$

$\to cos (x) + \frac{{sin}^{2} (x)}{cos (x)} = 2$ $rarr cos(x) + sin^2(x)/cos(x) = 2$

$\to \frac{{cos}^{2} (x) + {sin}^{2} (x)}{cos (x)} = 2$ $rarr (cos^2(x) + sin^2(x))/cos(x) = 2$

$\to \frac{1}{cos (x)} = 2$ $rarr 1/cos(x) = 2$

$\to cos (x) = \frac{1}{2}$ $rarr cos(x) = 1/2$

Based on one of the standard trigonometric triangles
$x = \pm 60^{\circ}$ $x = +-60^@$

How do you solve cosx+sinxtanx=2cosx+sinxtanx=2cosx+sinxtanx=2?