Question

9%20as%20%5C%3A%20%5C%3A%20...%5Cgamma%20%20%5Csin%282x%20-%20%20%5Calpha%20%29.%20%20%5Calpha%20%5C%3A%20%20is%20%5C%3A%20%20%5C%5C%20%20given%20%5C%3A%20%5C%3A%20%20as%20%5C%3A%20%20%5C%3A%20%20%280%20%3C%20%20%5Calpha%20%20%3C%20%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%29%20%5C%3A%20and%20%5C%3A%20%20%5C%3A%20%20%5Cgamma%20%28%20%3E%200%29.%20%5C%3A%20find%20%5C%3A%20%20%5C%3A%20%20%5Calpha%20%5C%3A%20%20%5C%3A%20%20and%20%20%5C%3A%20%20%20%5C%3A%20%5Cgamma%20...%20%5C%5C%20%20%5C%5C%20find%20%5C%3A%20the%20%5C%3A%20solutions%20%5C%3A%20for%20%5C%3A%20%20%5Csin%282x%29%20%20-%20%20%5Csqrt%7B3%7D%20%20%5Ccos%282x%29%20%20%3D%201%20%5C%3A%20between%20...%5C%3A%20%20-%20%5Cpi%20%5Cleqslant%20x%20%5Cleqslant%20%5Cpi" id="TexFormula1" title=" 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" alt=" 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" align="absmiddle" class="latex-formula">
can someone help me with this plzz​

Answer 1

Answer:

α = π/3

γ = 2

x = -3π/4, -5π/12, π/4, 7π/12

Step-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 = 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