Question

`express \: \: \sin(2x) - √(3) \cos(2x) as \: \: ...\gamma \sin(2x - \alpha ). \alpha \: is \: \\ given \: \: as \: \: (0 < \alpha < (\pi)/(2) ) \: and \: \: \gamma ( > 0). \: find \: \: \alpha \: \: and \: \: \gamma ... \\ \\ find \: the \: solutions \: for \: \sin(2x) - √(3) \cos(2x) = 1 \: between ...\: - \pi \leqslant x \leqslant \pi`

can someone help me with this plzz​

Answer 1

α = π/3

γ = 2

x = -3π/4, -5π/12, π/4, 7π/12

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 = 1

γ sin α = √3

Solve the system of equations. Divide to eliminate γ:

tan α = √3

α is between 0 and π/2, so:

α = π/3

γ = 2

sin(2x) − √3 cos(2x) = 1

Using the identity from before:

2 sin(2x − π/3) = 1

Solving:

sin(2x − π/3) = 1/2

2x − π/3 = π/6 + 2kπ or 5π/6 + 2kπ

2x = π/2 + 2kπ or 7π/6 + 2kπ

x = π/4 + kπ or 7π/12 + kπ

x is between -π and π, so:

x = -3π/4, -5π/12, π/4, 7π/12