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How would I solve cos x * cos 2x=0 ?

Answer 1

Violet Anderson

#x=2npi+-pi/2# or #x=npi+-pi/4#

#cosx(cos2x)=0# #=>" either " cosx=0=>x=2npi+-pi/2 " or "# #cos2x=0=>2x=2npi+-pi/2" or "x=npi+-pi/4 #

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Answer 2

Victoria Adebayo

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

cos x.cos 2x = 0 Either factor should be zero. a. cos x = 0 Unit circle gives --> #x = pi/2 + 2kpi#, and #x = (3pi)/2 + 2kpi# b. cos 2x = 0 Unit circle gives --> 1. #2x = pi/2 + 2kpi# #x = pi/4 + kpi# 2. #2x = (3pi)/2 + 2kpi# #x = (3pi)/4 + kpi#

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Answer 3

Nolan Alexander

To solve cos(x) * cos(2x) = 0, you need to find the values of x that make the product equal to zero. This occurs when either cos(x) = 0 or cos(2x) = 0. For cos(x) = 0, the solutions are x = π/2 + πn and x = 3π/2 + πn, where n is an integer. For cos(2x) = 0, the solutions are 2x = π/2 + πn or 2x = 3π/2 + πn, where n is an integer. Solving for x, you get x = π/4 + πn or x = 3π/4 + πn, where n is an integer.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.