Question

The general solution(s) of the equation $4{\cos }^{2}x+6{\sin }^{2}x=5$ is/are

• A. $x=n\pi +\frac{\pi }{2},n\in Z.$
• B. $x=n\pi -\frac{\pi }{2},n\in Z.$
• C. $x=n\pi -\frac{\pi }{4},n\in Z.$
• D. $x=n\pi +\frac{\pi }{4},n\in Z.$

Answer 1

Answer: (C), (D)

$x=n\pi +\frac{\pi }{4},n\in Z.$

Here, we can write the equation as:
$4{\cos }^{2}x+6{\sin }^{2}x=5\Rightarrow 4(1-{\sin }^{2}x)+6{\sin }^{2}x=5\Rightarrow 4-4{\sin }^{2}x+6{\sin }^{2}x=5\Rightarrow 2{\sin }^{2}x=1\Rightarrow {\sin }^{2}x=\frac{1}{2}={\left(\frac{1}{\sqrt{2}}\right)}^2={\sin }^2\frac{\pi }{4}.\mathrm{Thus},\mathrm{solution}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}:x=n\pi \pm \frac{\pi }{4},n\in Z.$