Answered

Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

Solve sin(x) = 0.23 on 0 ≤ x < 2T.There are two solutions, A and B, with A

Solve Sinx 023 On 0 X Lt 2TThere Are Two Solutions A And B With A class=

Sagot :

Given -

sin(x) = 0.23 on 0 ≤ x < 2π

To Find -

Two solutions, A and B =?

Step-by-Step Explanation -

[tex]0.23\text{ = }\frac{23}{100}[/tex]

So, the two solutions will be -

[tex]\begin{gathered} \sin^{-1}(\frac{23}{100})\text{ and }\pi\text{ - }\sin^{-1}(\frac{23}{100}) \\ \\ So,\text{ }\sin^{-1}(\frac{23}{100})\text{ = 0.232} \\ \\ Also,\text{ }\pi\text{ - }\sin^{-1}(\frac{23}{100})\text{ = 3.141 - 0.232 = 2.909} \end{gathered}[/tex]

Now, Since A < B

So,

A = 0.232

B = 2.909

Final Answer -

A = 0.232

B = 2.909

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.