End of school level equation

[Guest]

I don’t understand how to solve it and in what sequence?

[Guest]

sin(2x)= 2 sin(x)cos(x) and cos(2x)= cos^2(x)- sin^2(x)
so this equation is
$2\sqrt{3}sin(x)cos(x)- cos^2(x)+ sin^2(x)+ 2\sqrt{3} sin(x)+ 2 cos(x)= a$

We can write that as
$sin^2(x)+ 2\sqrt{3}(cos(x)+ 1)sin(x)+ (2cos(x)- cos^2(x)- a)= 0$

Use the quadrtic formula- the solutions to $ax^2+ bx+ c= 0$ are $x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$-
with $a= 1$, $b= 2\sqrt{3}(cos(x)+ 1)$, and $c= 2cos(x)- cos^2(x)- a$ to get an equation relating sin(x) and cos(x) and solve that equation.

[Guest]

Thanks!

[zoryn]

I don’t understand how to solve it and in what sequence?

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yeah try rewriting it with the double angle formulas first
√3 * 2sinx cosx - (cos²x - sin²x) + 2√3 sinx + 2 cosx = a
then group the terms and treat it like a quadratic in sinx or shift the angle to get something like 4sin²(x+π/6) + 4sin(x+π/6) = a+2
that usually works for these end of school trig eqs