How do you solve #cos(2x)+cos(2x)*sin(2x)=0# in the interval [0, 2pi]?

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How do you solve #cos(2x)+cos(2x)*sin(2x)=0# in the interval [0, 2pi]?

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Answer 1 · 2016-02-07T05:09:00

#x=pi/2,pi/4, 3pi/4#

Explanation:

let # f(x)= cos(2x) + cos(2x)sin(2x) #
Solve for x in # f(x)= 0#

# cos(2x) + cos(2x)sin(2x) = 0 #
# cos(2x)[1+sin(2x)] = 0 #
in there equation , the only way #f(x)# wil be #0#
if the multiplier of the multiplicand is 0

so # cos(2x)=0 #
for cos#(2x)# to be #0#
#2x=pi/2, 3pi/2#
#x = pi/4, 3pi/4#

Or # 1+sin(2x)=0#
#sin(2x)=-1#
# 2x= pi #
#x=pi/2#

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How do you solve #cos(2x)+cos(2x)*sin(2x)=0# in the interval [0, 2pi]?

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