[tex] express \: \: \sin(2x) - \sqrt{3} \cos(2x) as \: \: ...\gamma \sin(2x - \alpha ). \alpha \: is \: \\ given \: \: as \: \: (0 < \alpha < \frac{\pi}{2} ) \: and \: \: \gamma ( > 0). \: find \: \: \alpha \: \: and \: \: \gamma ... \\ \\ find \: the \: solutions \: for \: \sin(2x) - \sqrt{3} \cos(2x) = 1 \: between ...\: - \pi \leqslant x \leqslant \pi[/tex]can someone help me with this plzz​

Answer:α = π/3γ = 2x = -3π/4, -5π/12, π/4, 7π/12Step-by-step explanation:Easiest way to do this is in reverse.γ sin(2x − α)Angle sum/difference formula:γ (cos α sin(2x) − sin α cos(2x))Distribute:γ cos α sin(2x) − γ sin α cos(2x)Matching the coefficients:γ cos α = 1γ sin α = √3Solve the system of equations.  Divide to eliminate γ:tan α = √3α is between 0 and π/2, so:α = π/3γ = 2sin(2x) − √3 cos(2x) = 1Using the identity from before:2 sin(2x − π/3) = 1Solving:sin(2x − π/3) = 1/22x − π/3 = π/6 + 2kπ or 5π/6 + 2kπ2x = π/2 + 2kπ or 7π/6 + 2kπx = π/4 + kπ or 7π/12 + kπx is between -π and π, so:x = -3π/4, -5π/12, π/4, 7π/12