Question #af5d8

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Question #af5d8

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Answer 1 · 2017-05-02T13:29:50

a. cos 2x = 0
Unit circle gives:
$2 x = \frac{π}{2} + 2 k π$ $2x = pi/2 + 2kpi$ and $2 x = \frac{3 π}{2} + 2 k π$ $2x = (3pi)/2 + 2kpi$

$2 x = \frac{π}{2} + 2 k π$ $2x = pi/2 + 2kpi$ --> $x = \frac{π}{4} + k π$ $x = pi/4 + kpi$
$2 x = \frac{3 π}{2} + 2 k π$ $2x = (3pi)/2 + 2kpi$ --> $x = \frac{3 π}{4} + k π$ $x = (3pi)/4 + kpi$
Answers for $(0, 2 π)$ $(0, 2pi)$ :
$\frac{π}{4}; \frac{3 π}{4}; \frac{5 π}{4}; \frac{7 π}{4}$ $pi/4; (3pi)/4; (5pi)/4; (7pi)/4$

b. $cos x = cos 2 x = 2 {cos}^{2} x - 1$ $cos x = cos 2x = 2cos ^2 x - 1$
Solve this quadratic equation for cos x:
$2 {cos}^{2} x - cos x - 1 = 0$ $2cos^2 x - cos x - 1 = 0$
Since a + b + c = 0, use shortcut.
The 2 real roots are: cos x = 1 and cos x = c/a = - 1/2
Trig table and unit circle give:
1/ cos x = 1 --> x = 0 and x = 2kpi
2/ $cos x = - \frac{1}{2}$ $cos x = - 1/2$ --> $x = \pm \frac{2 π}{3} + 2 k π$ $x = +- (2pi)/3 + 2kpi$
Answers for $(0, 2 π)$ $(0, 2pi)$ :
$2 π; \pm \frac{2 π}{3}$ $2pi; +- (2pi)/3$
c. sin 3x = 0
Unit circle gives as solutions:
1/ $3 x = 0 + 2 k π$ $3x = 0 + 2kpi$ --> $x = \frac{2 k π}{3}$ $x = (2kpi)/3$
2/ $3 x = π + 2 k π$ $3x = pi + 2kpi$ --> $x = \frac{(2 k + 1) π}{3}$ $x = ((2k + 1)pi)/3$
3/ 3x = 2pi + 2kpi --> x = (2kpi)/3
Answers for $(0, 2 π)$ $(0, 2pi)$ :
$\frac{2 π}{3}; \frac{4 π}{3}; 2 π; \frac{π}{3}; π$ $(2pi)/3; (4pi)/3; 2pi; pi/3; pi$

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Question #af5d8

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