Question #7fce1

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Question #7fce1

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Answer 1 · 2017-12-07T06:02:51

$x = \frac{π}{4} + 2 π k, \frac{π}{2} + 2 π k, 3 \frac{π}{4} + 2 π k$ $x = pi/4 + 2pik, pi/2 + 2pik, 3pi/4 + 2pik$ (Period of sin(x) and cos(x) is $2 π$ $2pi$ )

Explanation:

Assuming $sin (2 x) sin (x) = cos (x)$ $sin(2x)sin(x) = cos(x)$ , begin by using the double angle identity, $sin (2 u) = 2 sin (u) cos (u)$ $sin(2u) = 2sin(u)cos(u)$ to rewrite the equation.

$sin (2 x) sin (x) = cos (x) \to 2 sin (x) sin (x) cos (x) = cos (x) \to 2 {sin}^{2} (x) cos (x) = cos (x)$ $sin(2x)sin(x) = cos(x) rarr 2sin(x)sin(x)cos(x) = cos(x) rarr 2sin^2(x)cos(x) = cos(x)$

Next subtract cos(x) from the right side to the left to make the equation equal to zero.

$2 {sin}^{2} (x) cos (x) - cos (x) = 0$ $2sin^2(x)cos(x)-cos(x) = 0$

Factor out cos(x) and separate

$cos (x) \cdot (2 {sin}^{2} (x) - 1) = 0 \to cos (x) = 0$ $cos(x)*(2sin^2(x)-1) = 0 rarr cos(x) = 0$ and $2 {sin}^{2} (x) - 1 = 0$ $2sin^2(x)-1 = 0$

Find the exact values using the unit circle or right triangle method, $cos (x) = 0$ $cos(x) = 0$ when $x = \frac{π}{2}$ $x = pi/2$ and $sin (x) = \frac{\sqrt{2}}{2}$ $sin(x) = sqrt(2)/2$ when $x = \frac{π}{4}, 3 \frac{π}{4}$ $x = pi/4, 3pi/4$ (Remember restrictions $sin (x) : R (- \frac{π}{2}, \frac{π}{2}); cos (x) : R (0, π)$ $sin(x): R(-pi/2, pi/2); cos(x): R(0, pi)$

Answer 2 · 2017-12-07T17:23:00

Answers:
$x = \frac{π}{2} + k π$ $x = pi/2 + kpi$
$x = \frac{π}{4} + \frac{k π}{2}$ $x = pi/4 + (kpi)/2$

Explanation:

$sin 2 x . sin x - cos x = 0$ $sin 2x.sin x - cos x = 0$
Replace sin 2x by (2sin x.cos x).
$2 {sin}^{2} x . cos x - cos x = 0$ $2sin^2 x.cos x - cos x = 0$
$cos x (2 {sin}^{2} x - 1) = 0$ $cos x(2sin^2 x - 1) = 0$
Either factor should be zero.
A. cos x = 0. Unit circle give 2 solutions:
$x = \frac{π}{2},$ $x = pi/2,$ and $x = \frac{3 π}{2}$ $x = (3pi)/2$
General answer: $x = \frac{π}{2} + k π$ $x = pi/2 + kpi$
B. $2 {sin}^{2} x - 1 = 0$ $2sin^2 x - 1 = 0$ --> ${sin}^{2} x = \frac{1}{2}$ $sin^2 x = 1/2$
$sin x = \pm \frac{\sqrt{2}}{2}$ $sin x = +- sqrt2/2$
a. $sin x = \frac{\sqrt{2}}{2}$ $sin x = sqrt2/2$ . Trig table and unit circle give 2 solutions:
$x = \frac{π}{4}$ $x = pi/4$ , and $x = π - \frac{π}{4} = \frac{3 π}{4}$ $x = pi - pi/4 = (3pi)/4$
b. $sin x = - \frac{\sqrt{2}}{2}$ $sin x = -sqrt2/2$ . Unit circle gives -->
$x = π + \frac{π}{4} = \frac{5 π}{4}$ $x = pi + pi/4 = (5pi)/4$ and $x = 2 π - \frac{π}{4} = \frac{7 π}{4}$ $x = 2pi - pi/4 = (7pi)/4$
General answer: $x = \frac{π}{4} + \frac{k π}{2}$ $x = pi/4 + (kpi)/2$

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