Answer:
2. [tex]x = \frac{\pi }{4} , \frac{3\pi }{4}, \frac{5\pi }{4}, \frac{7\pi }{4}[/tex]
3. -0.28
4. a. [tex]- \frac{\pi }{6}[/tex]
b. [tex]\frac{3\pi }{4}[/tex]
Step-by-step explanation:
2. Given that
[tex]2\sin^{2}x = 1[/tex]
⇒ [tex]\sin x = \frac{1}{\sqrt{2} }[/tex] or [tex]\sin x = - \frac{1}{\sqrt{2} }[/tex]
Since x belongs to [0,2π).
Therefore, [tex]x = \frac{\pi }{4} , \frac{3\pi }{4}, \frac{5\pi }{4}, \frac{7\pi }{4}[/tex] (Answer)
3. Given that, [tex]\cos \theta = \frac{3}{5}[/tex]
⇒ [tex]\theta = \cos^{-1}(\frac{3}{5}) = 53.13[/tex] Degrees
{Since [tex]\theta[/tex] is in first quadrant}
So, [tex]\cos 2\theta = \cos (2 \times 53.13) = -0.28[/tex] (Answer)
4. a. Given [tex]\sin^{-1}(- \frac{1}{2}) = - \frac{\pi }{6}[/tex]
b. [tex]\cos^{-1}(- \frac{\sqrt{2} }{2} ) = \cos^{-1}(- \frac{1}{\sqrt{2} } ) = \frac{3\pi }{4}[/tex] (Answer)
