How do you solve #cos x tan x = 1/2# on the interval #[0,2pi]#?

1 Answer
Daan J.
Jul 17, 2015

#x=pi/6#, or #x=5pi/6#

Explanation:

We note that #tanx=sinx/cosx#, so #cosxtanx=1/2# is equivalent to #sinx=1/2#, this gives us #x=pi/6#, or #x=5pi/6#. We can see this, using the fact that if the hypotenuse of a right triangle is twice the size of the opposite side of one of the non-right angles, we know that the triangle is half an equilateral triangle, so the inner angle is half of #60^@=pi/3 "rad"#, so #30^@=pi/6 "rad"#. We also note that the outer angle (#pi-pi/6=5pi/6#) has the same value for its sine as the inner angle. Since this is the only triangle where this occurs, we know these solutions are the only two possible solutions on the interval #[0,2pi]#.

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