How do you solve $cos x = sin 2 x sin x$ $cosx=sin2xsinx$ ?

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How do you solve $cos x = sin 2 x sin x$ $cosx=sin2xsinx$ ?

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Answer 1 · 2016-09-12T04:24:06

$x = \frac{π}{4} or 45^{\circ}$ $x=pi/4 or 45^@$

Explanation:

recall that $sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$

$cos x = sin 2 x sin x$ $cosx=sin2xsinx$
$cos x = 2 sin x cos x sin x$ $cosx=2sinxcosxsinx$
$1 = 2 {sin}^{2} x$ $1=2sin^2x$
${sin}^{2} x = \frac{1}{2}$ $sin^2x=1/2$

$sin x = \frac{1}{\sqrt{2}}$ $sinx=1/sqrt2$

$x = 45^{\circ}$ $x=45^@$

Answer 2 · 2016-09-12T12:52:01

$\frac{π}{2}; \frac{3 π}{2}; \frac{π}{4}; \frac{3 π}{4}; \frac{5 π}{4}; \frac{7 π}{4}$ $pi/2; (3pi)/2; pi/4; (3pi)/4; (5pi)/4; (7pi)/4$

Explanation:

cos x = sin 2x.sin x
$cos x = 2 sin x . cos x . sin x = 2 cos x . {sin}^{2} x$ $cos x = 2sin x.cos x.sin x = 2cos x.sin^2 x$
$cos x - 2 cos x . {sin}^{2} x = 0$ $cos x - 2cos x.sin^2 x = 0$
Put cos x in common factor:
$cos x (1 - 2 {sin}^{2} x) = 0$ $cos x(1 - 2sin^2 x) = 0$
Use trig table, and unit circle to solve this trig equation
a. cos x = 0 --> $x = \frac{π}{2} and x = \frac{3 π}{2}$ $x = pi/2 and x = (3pi)/2$
b. $(1 - 2 {sin}^{2} x) = 0$ $(1 - 2sin^2 x) = 0$
${sin}^{2} x = \frac{1}{2}$ $sin^2 x = 1/2$
$sin x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$ $sin x = +- 1/sqrt2 = +- sqrt2/2$
c. $sin x = \frac{\sqrt{2}}{2}$ $sin x = sqrt2/2$ --> $x = \frac{π}{4} and x = \frac{3 π}{4}$ $x = pi/4 and x = (3pi)/4$
d. $sin x = - \frac{\sqrt{2}}{2}$ $sin x = - sqrt2/2$ --> $x = \frac{5 π}{4} and x = \frac{7 π}{4}$ $x = (5pi)/4 and x = (7pi)/4$

Answers for $(0, 2 π)$ $(0, 2pi)$ :
$\frac{π}{2}; \frac{3 π}{2}; \frac{π}{4}; \frac{3 π}{4}; \frac{5 π}{5}; \frac{7 π}{4} .$ $pi/2; (3pi)/2; pi/4; (3pi)/4; (5pi)/5; (7pi)/4.$
For general answers, add $2 k π$ $2kpi$

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