How do you solve $2 {cos}^{2} (x) + 4 {sin}^{2} (x) = 3$ $2cos^2(x)+ 4sin^2(x)=3$ ?

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Answer 1 · 2015-05-11T19:41:35

Using ${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2(x)+cos^2(x) = 1$ , we can express the equation as:

$3 = 2 {cos}^{2} (x) + 4 {sin}^{2} (x) = 2 (1 - {sin}^{2} (x)) + 4 {sin}^{2} (x)$ $3 = 2cos^2(x)+4sin^2(x) = 2(1-sin^2(x))+4sin^2(x)$

$= 2 - 2 {sin}^{2} (x) + 4 {sin}^{2} (x)$ $=2-2sin^2(x)+4sin^2(x)$

$= 2 + 2 {sin}^{2} (x)$ $=2+2sin^2(x)$

Subtract 2 from both ends to get:

$2 {sin}^{2} (x) = 1$ $2sin^2(x)=1$

Divide both sides by 2 to get:

${sin}^{2} (x) = \frac{1}{2}$ $sin^2(x)=1/2$

Again, since ${sin}^{2} (x) + {cos}^{2} (x) = 1$ $sin^2(x)+cos^2(x) = 1$ , we also have:

${cos}^{2} (x) = 1 - {sin}^{2} (x) = 1 - \frac{1}{2} = \frac{1}{2}$ $cos^2(x)=1-sin^2(x)=1-1/2=1/2$

Hence $sin (x) = \pm \frac{\sqrt{2}}{2}$ $sin(x)=+-sqrt(2)/2$ and $cos (x) = \pm \frac{\sqrt{2}}{2}$ $cos(x)=+-sqrt(2)/2$

This is true for $x = \frac{π}{4} + \frac{n π}{2}$ $x = pi/4+(n pi)/2$ for all integer values of $n$ $n$ .

How do you solve 2cos^2(x)+ 4sin^2(x)=32cos2(x)+4sin2(x)=32cos^2(x)+ 4sin^2(x)=3?