Thank you very much for the points! Helped me a lot!
as you already know, to get the inverse of any expression we start off by doing a quick switcheroo on the variables and then solving for "y", let's do so.
![\stackrel{h(x)}{y}~~ = ~~6\sqrt[3]{2x+5}-1\implies \stackrel{\textit{quick switcheroo}}{x~~ = ~~6\sqrt[3]{2y+5}-1} \\\\\\ x+1=6\sqrt[3]{2y+5}\implies \cfrac{x+1}{6}=\sqrt[3]{2y+5}\implies \left( \cfrac{x+1}{6} \right)^3=\left( \sqrt[3]{2y+5} \right)^3](https://tex.z-dn.net/?f=%5Cstackrel%7Bh%28x%29%7D%7By%7D~~%20%3D%20~~6%5Csqrt%5B3%5D%7B2x%2B5%7D-1%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bquick%20switcheroo%7D%7D%7Bx~~%20%3D%20~~6%5Csqrt%5B3%5D%7B2y%2B5%7D-1%7D%20%5C%5C%5C%5C%5C%5C%20x%2B1%3D6%5Csqrt%5B3%5D%7B2y%2B5%7D%5Cimplies%20%5Ccfrac%7Bx%2B1%7D%7B6%7D%3D%5Csqrt%5B3%5D%7B2y%2B5%7D%5Cimplies%20%5Cleft%28%20%5Ccfrac%7Bx%2B1%7D%7B6%7D%20%5Cright%29%5E3%3D%5Cleft%28%20%5Csqrt%5B3%5D%7B2y%2B5%7D%20%5Cright%29%5E3)

Answer:
Step-by-step explanation:
Answer:
x∈[2nπ−5π/6, 2nπ−π/6]∪{(4n+1)π/2}, n ϵ I
Step-by-step explanation:
1−2sin2 x≤−sin x ⇒ (2sin x+1)(sin x−1)≥0
sin x≤−1/2 or sin x≥1
−5π/6+2nπ≤x≤−π/6+2nπ or , n ϵ I x=(4n+1)π/2, n ϵ I⇒ -5π6+2nπ≤x≤-π6+2nπ or , n ϵ I x=4n+1π2, n ϵ I (as sin x = 1 is valid only)
In general⇒ In general x∈[2nπ−5π/6, 2nπ−π/6]∪{(4n+1)π/2}, n ϵ I
Answer:
Step-by-step explanation:
You need to add more description, Just by saying this question doesnt really help the reader understand what you are trying to say.