Question

solve sin(3x + 5π/4) =1 for -π≤x≤π

Answer 1

\[\text{ let } \theta=3x+\frac{5 \pi}{4} \]
Solve for x so we can find the inequality in terms of theta.
\[\theta-\frac{5 \pi}{4}= 3x \\ 3x=\theta-\frac{ 5 \pi }{4} \\ x=\frac{ \theta}{3} -\frac{ 5 \pi}{3(4)} \\ x =\frac{ \theta}{3} -\frac{5 \pi}{12 } \\ \text{ so we have } -\pi \le x \le \pi \\ \text{ writing in terms of } \theta \\ -\pi \le \frac{ \theta}{3}-\frac{5 \pi}{12} \le \pi\]
now solve the inequality for theta...
\[\text{ multiply both sides by } 12 \\ -12 \pi \le 4 \theta - 5\pi \le 12 \pi \\ \text{ add } 5\pi \text{ on both sides } \\ -7 \pi \le 4 \theta \le 17 \pi \\ \text{ divide both sides by } 4 \\ \frac{ - 7 \pi}{4} \le \theta \le \frac{ 17 \pi}{4} \\ \text{ so the first to solve your equation } \\ \text{ is to solve this equation on the given interval for } \theta \\ \sin(\theta)=1 \text{ on } \theta \in [\frac{ -7 \pi}{4} , \frac{ 17 \pi}{4}]\]
then once you find the solution for theta
we will replace theta with the 3x+5pi/4 thing and solve for x
(don't worry this will just be a linear equation in the end)