Question

how to solve sin(2x)=-cos(2x)

Answer 1

\[
\frac{\sin(2x)}{\cos(2x)} = \frac{-1}{1}
\]

Answer 2

Or \(\tan(2x) = -1\).
This means \(2x=3\pi/4\) or \(2x=7\pi/4\).

Answer 3

i need to solve x whose domain is [-3∏/2, 3∏/2] and there should be 6 answers

Answer 4

[-3π/2, 3π/2]

Answer 5

\[\tan(2x)=-1 , \frac{-3\pi}{2} \le x \le \frac{3\pi}{2} \\ \text{ let } u=2x\text{ then } \frac{u}{2}=x \\ \text{ so we have solve for u: } \\ \tan(u)=-1, \frac{-3\pi}{2} \le \frac{u}{2} \le \frac{3\pi}{2} \\ \text{ rewrite } \\ \text{ solve }\tan(u)=-1, -3\pi \le u \le 3\pi \]
find all u that satisfy tan(u)=-1 in the interval [-3pi,3pi]
then replace u with 2x
and solve for x

Answer 6

thank you bro that makes sence