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

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Answer 1 · 2015-05-11T19:41:35

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

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

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

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

Subtract 2 from both ends to get:

$2sin^2(x)=1$

Divide both sides by 2 to get:

$sin^2(x)=1/2$

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

$cos^2(x)=1-sin^2(x)=1-1/2=1/2$

Hence $sin(x)=+-sqrt(2)/2$ and $cos(x)=+-sqrt(2)/2$

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

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