How do you solve ${sin}^{2} (2 x) + {cos}^{2} x = 0$ $sin^2(2x)+cos^2x=0$ ?

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How do you solve ${sin}^{2} (2 x) + {cos}^{2} x = 0$ $sin^2(2x)+cos^2x=0$ ?

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Answer 1 · 2017-02-23T04:09:23

For (x, 2pi)
$\frac{π}{6}; \frac{π}{2}; \frac{5 π}{6}; \frac{7 π}{6}; \frac{3 π}{2}; \frac{11 π}{6}$ $pi/6; pi/2; (5pi)/6; (7pi)/6; (3pi)/2; (11pi)/6$

Explanation:

Reminder:
${sin}^{2} (2 x) = 4 {sin}^{2} x . {cos}^{2} x$ $sin^2 (2x) = 4sin^2 x.cos^2 x$
We get:
${sin}^{2} (2 x) + {cos}^{2} x = {cos}^{2} x (1 + 4 {sin}^{2} x) = 0$ $sin^2 (2x) + cos^2 x = cos^2 x(1 + 4sin^2 x) = 0$
a. ${cos}^{2} x = 0$ $cos^2 x = 0$ --> cos x = 0 --> 2 solutions:
$x = \frac{π}{2}$ $x = pi/2$ and $x = \frac{3 π}{2}$ $x = (3pi)/2$
b. $1 + 4 {sin}^{2} x = 0$ $1 + 4sin^2 x = 0$ --> ${sin}^{2} x = \frac{1}{4}$ $sin^2 x = 1/4$ -->
$sin x = \pm \frac{1}{2}$ $sin x = +- 1/2$
Use trig table and unit circle.
$sin x = \frac{1}{2}$ $sin x = 1/2$ --> $x = \frac{π}{6}$ $x = pi/6$ and $x = \frac{5 π}{6}$ $x = (5pi)/6$
$sin x = - \frac{1}{2}$ $sin x = - 1/2$ --> $x = \frac{7 π}{6}$ $x = (7pi)/6$ and $x = \frac{11 π}{6}$ $x = (11pi)/6$

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How do you solve ${sin}^{2} (2 x) + {cos}^{2} x = 0$ $sin^2(2x)+cos^2x=0$ ?

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