Please answer thoroughly. If 2sin(x)-cos(x)=(5/12), solve for the angle x.?

2*sin(x) = cos(x) + 5/12 =>

4 sin^2(x) = cos^2(x) + (5/6) cos(x) + 25/144 =>

4(1 -cos^2(x)) = cos^2(x)+ (5/6)cos(x) +25/144=>

0 = 5 cos^2(x) + (5/6) cos(x) - 551/144.

Let u = cos(x), you have

0 = 5u^2 + (5/6) u - 551/144.

u = (-1/12) +/- (1/10)*sqrt(25/36 + 2755/36)

= (-1/12) +/- (1/10)*sqrt(2780/36)

= (-1/12) +/- (1/10)*sqrt(695/9)

= (-1/12) +/- 0.87876

= 0.79543 or -0.96209.

If I haven't made a mistake yet, now just use inverse cosine function, answers seem to be around 37.3 degrees and 164.2 degrees.

Wolframalpha will give an answer. It's ugly.

Sin(x)^2+cos(x)^2=1

so cos(x)=radical(1-sin(x)^2)

Now lets rewrite your equation using radical(1-sin(x)^2) instead of cos(x)

2sin(x)-radical(1-sin(x)^2) =(5/12)

So 2sin(x)-(5/12)=radical(1-sin(x)^2)

Raise both sides of the equation to power 2

4sin(x)^2+25/144-5/3sin(x)=1-sin(x)^2

If u solve this equation u will see that sin(x) is either 5/6 or 15/6 in which the latter is not acceptable(sin(x) cant be bigger than 1) so angle x is arcsin(5/6)