Question 1195997
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I read this equation in this way

    {{{cos(2*alpha + pi/4)}}} = {{{3*tan(2*alpha + pi/4)}}}.


Notice that the argument under cosine function is the same as under the tangent function.

So, I introduce new variable

    x = {{{2*alpha + pi/4}}}.


Then the given equation takes the form

    cos(x) = 3*tan(x).


It implies

    cos^2(x) = 3*sin(x),

    1-sin^2(x) = 3*sin(x)

    sin^2(x) + 3*sin(x) - 1 = 0

It is a quadratic equation relative sin(x).  Use the quadratic formula

    sin(x) = {{{(-3 +- sqrt(3^2-4*1*(-1)))/2}}} = {{{(-3 +- sqrt(13))/2}}}.


Since sin(x) must be between -1 and 1, only the root sin(x) = {{{(-3 + sqrt(13))/2}}} = 0.30278  really fits.


Hence,  there are two solutions for x:

    x = arcsin(0.30278) = 0.3076  OR  x = {{{pi-arcsin(0.30278)}}} = 3.14159 - 0.3076 = 2.834 (rounded).


It gives two solutions for {{{alpha}}}:

    (a)  {{{alpha}}} = {{{(1/2)*(0.3076 - pi/4)}}} = {{{(1/2)*(0.3076-3.14159/4)}}} = -0.2389  (rounded);

    (b)  {{{alpha}}} = {{{(1/2)*(2.834 - pi/4)}}}  = {{{(1/2)*(2.834-3.14159/4)}}}  =  1.0243  (rounded).
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Solved.