Question #92a25

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Question #92a25

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Answer 1 · 2017-04-03T07:25:00

$x=0,pi/2,2pi$

Explanation:

$1-sinx=cosx$

$=>cosx+sinx=1$

$=>1/sqrt2cosx+1/sqrt2sinx=1/sqrt2$

$=>cos(pi/4)cosx+sin(pi/4)sinx=cos(pi/4)$

$=>cos(x-pi/4)=cos(pi/4)=cos(-pi/4)=cos(2pi-pi/4)$

So $x-pi/4=pi/4$

$=>x=pi/2$

when

$x-pi/4=-pi/4$

$x=0$

Again

$x-pi/4=2pi-pi/4$

$=>x=2pi$

Answer 2 · 2017-04-04T00:48:08

$x = 2kpi$
$x = pi/2 + 2kpi$

Explanation:

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

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Question #92a25

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