Finding the required angle in the triangle

I have the following question with me:

"Consider a triangle $$ABC$$ and let $$M$$ be the midpoint of the side $$BC$$. Suppose $$\angle MAC$$ = $$\angle ABC$$ and $$\angle BAM = 105^{\circ}$$ . Find the measure of $$∠ABC$$."

Upon taking $$\angle ABC$$ as $$x$$ and upon simplifying, I get the following equation, if my calculations are correct,

$$(4+\sqrt{3})\cos2x = 2 + \sin2x$$

How do I solve this equation?

Alternative solution.

Let $$BC = 2a$$. By the power of the point of $$C$$ with respect to circle $$(ABM)$$, which is tangent to a line $$AC$$, we have $$CA^2= CM\cdot CB =2a^2$$

By rule of sine for $$ABC$$ we have $${2a\over \sin (\alpha +105^{\circ})} ={a\sqrt{2}\over \sin \alpha}$$

so $$\cot \alpha = {\sqrt{2}+\sin 15^{\circ}\over \cos 15^{\circ}} =\sqrt{3}\implies \alpha = 30^{\circ}$$

• How is $AC$ tangent to circumcircle of $ABM$? – saisanjeev Oct 2 '18 at 7:29
• Because $\angle MAC = \angle ABC$ and tangent chord property. – Aqua Oct 2 '18 at 7:30

You can either use t-method, letting $$t=\tan x$$ so that $$\sin 2x=\frac {2t}{1+t^2}$$ and $$cos 2x=\frac {1-t^2}{1+t^2}$$. Plugging this in and solving the quadratic gives you values of $$t$$ which you can these use to solve for $$x$$. Alternatively, you can use auxiliary angles where $$A \cos 2x-B \sin 2x=\sqrt{(A^2+B^2)\cos (2x+\arctan \frac {B}{A})}$$