Question #92a25

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Question #92a25

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Answer 1 · 2017-04-03T07:25:00

$x = 0, \frac{π}{2}, 2 π$ $x=0,pi/2,2pi$

Explanation:

$1 - sin x = cos x$ $1-sinx=cosx$

$\Rightarrow cos x + sin x = 1$ $=>cosx+sinx=1$

$\Rightarrow \frac{1}{\sqrt{2}} cos x + \frac{1}{\sqrt{2}} sin x = \frac{1}{\sqrt{2}}$ $=>1/sqrt2cosx+1/sqrt2sinx=1/sqrt2$

$\Rightarrow cos (\frac{π}{4}) cos x + sin (\frac{π}{4}) sin x = cos (\frac{π}{4})$ $=>cos(pi/4)cosx+sin(pi/4)sinx=cos(pi/4)$

$\Rightarrow cos (x - \frac{π}{4}) = cos (\frac{π}{4}) = cos (- \frac{π}{4}) = cos (2 π - \frac{π}{4})$ $=>cos(x-pi/4)=cos(pi/4)=cos(-pi/4)=cos(2pi-pi/4)$

So $x - \frac{π}{4} = \frac{π}{4}$ $x-pi/4=pi/4$

$\Rightarrow x = \frac{π}{2}$ $=>x=pi/2$

when

$x - \frac{π}{4} = - \frac{π}{4}$ $x-pi/4=-pi/4$

$x = 0$ $x=0$

Again

$x - \frac{π}{4} = 2 π - \frac{π}{4}$ $x-pi/4=2pi-pi/4$

$\Rightarrow x = 2 π$ $=>x=2pi$

Answer 2 · 2017-04-04T00:48:08

$x = 2 k π$ $x = 2kpi$
$x = \frac{π}{2} + 2 k π$ $x = pi/2 + 2kpi$

Explanation:

sin x + cos x = 1
Use trig identity:
$sin x + cos x = \sqrt{2} sin (x + \frac{π}{4})$ $sin x + cos x = sqrt2sin (x + pi/4)$
In this case:
$\sqrt{2} sin (x + \frac{π}{4}) = 1$ $sqrt2sin (x + pi/4) = 1$
$sin (x + \frac{π}{4}) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $sin (x + pi/4) = 1/sqrt2 = sqrt2/2$
Trig table and unit circle give 2 solutions -->
a. $(x + \frac{π}{4}) = \frac{π}{4} + 2 k π$ $(x + pi/4) = pi/4 + 2kpi$
$x = 0 + 2 k π$ $x = 0 + 2kpi$ , or $x = 2 k π$ $x = 2kpi$
b. $(x + \frac{π}{4}) = π - \frac{π}{4} + 2 k π = \frac{3 π}{4} + 2 k π$ $(x + pi/4) = pi - pi/4 + 2kpi = (3pi)/4 + 2kpi$
$x = \frac{3 π}{4} - \frac{π}{4} + 2 k π = \frac{π}{2} + 2 k π$ $x = (3pi)/4 - pi/4 + 2kpi = pi/2 + 2kpi$

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Question #92a25

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