How do you solve sin2x = √2 cosx, 0≤x≤ π?

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How do you solve sin2x = √2 cosx, 0≤x≤ π?

sin2x = #sqrt(2)# cosx, 0#<=#x#<=##pi#?

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Answer 1 · 2017-03-24T16:28:18

#pi/2; pi/4; (3pi)/4#

Explanation:

Use trig identity : sin 2x = 2sin x.cos x
Substitute in the equation sin 2x by 2sin x.cos x
#2sin x.cos x - sqrt2cos x = 0#
#cos x(2sin x - sqrt2) = 0#
Either one of the 2 factors should be zero:
a. cos x = 0 --> #x = pi/2#, and #x = (3pi)/2#
b. #2sin x - sqrt2 = 0#
#sin x = sqrt2/2#
Trig unit circle gives 2 solutions:
#x = pi/4 and x = (3pi)/4#
Answers for #(0, pi)#
#pi/2; pi/4; (3pi)/4#

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How do you solve sin2x = √2 cosx, 0≤x≤ π?

sin2x = #sqrt(2)# cosx, 0#<=#x#<=##pi#?

1 Answer

Explanation:

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