Question

Hello, I need help with all these problems. So, I would like someone can help me

Answer 1

Answer:

2.  $x = \frac{\pi }{4} , \frac{3\pi }{4}, \frac{5\pi }{4}, \frac{7\pi }{4}$

3. -0.28

4. a. $- \frac{\pi }{6}$

b. $\frac{3\pi }{4}$

Step-by-step explanation:

2. Given that  

$2\sin^{2}x = 1$

⇒ $\sin x = \frac{1}{\sqrt{2} }$ or  $\sin x = - \frac{1}{\sqrt{2} }$

Since x belongs to [0,2π).

Therefore, $x = \frac{\pi }{4} , \frac{3\pi }{4}, \frac{5\pi }{4}, \frac{7\pi }{4}$ (Answer)

3. Given that,  $\cos \theta = \frac{3}{5}$

⇒ $\theta = \cos^{-1}(\frac{3}{5}) = 53.13$ Degrees  

{Since $\theta$ is in first quadrant}

So, $\cos 2\theta = \cos (2 \times 53.13) = -0.28$ (Answer)

4. a. Given $\sin^{-1}(- \frac{1}{2}) = - \frac{\pi }{6}$

b. $\cos^{-1}(- \frac{\sqrt{2} }{2} ) = \cos^{-1}(- \frac{1}{\sqrt{2} } ) = \frac{3\pi }{4}$ (Answer)