How do you solve #3sin^(2)x - sinxcosx - 4cos^(2)x = 0# from [-pi, pi]?

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How do you solve #3sin^(2)x - sinxcosx - 4cos^(2)x = 0# from [-pi, pi]?

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Answer 1 · 2015-09-07T13:41:11

3#y^2# - y#sqrt# (1 - #y^2# ) - 4 (1 - #y^2# ) = 0, y = sinx
#=># 7#y^2# = y#sqrt# (1 - #y^2# ) #=># 7y = #sqrt#(1 - #y^2# )
#=># y = 1/5#sqrt #2 #=># x = #sin^- 1# (1/ 5 #sqrt# 2 )

Explanation:

Squaring 7y = #sqrt#(1 - #y^2# ), we get 49 #y^2# = 1 - #y^2# #=>#
50 #y^2# = 1 #=># #y^2# = 1/50 #=># y = 1/5#sqrt #2 .

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How do you solve #3sin^(2)x - sinxcosx - 4cos^(2)x = 0# from [-pi, pi]?

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