How do you find the general solutions for #sqrt{3}(sin(x)) + cos(x) = 1#?

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How do you find the general solutions for #sqrt{3}(sin(x)) + cos(x) = 1#?

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Answer 1 · 2015-08-12T13:55:40

# x = (6n+(-1)^n-1)pi/6, " " n in ZZ #

Explanation:

# R = sqrt(a^2+b^2) #
# a sin x + b cos x = R(a/R sin x + b/R cos x) = R sin (x + alpha) #
where #a/R = cos alpha# and #b/R = sin alpha# such that # alpha = arctan (b/a) #

# sqrt(3) sin x + cos x = 2 sin (x + pi/6) = 1 #
# sin (x + pi/6) = 1/2 #

# x + pi/6 = npi + (-1)^npi/6, " " n in ZZ #
# x = (6n+(-1)^n-1)pi/6, " " n in ZZ #

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How do you find the general solutions for #sqrt{3}(sin(x)) + cos(x) = 1#?

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