How do you solve #sqrt2tanx=2sinx#?

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Answer 1 · 2017-01-08T18:02:52

Please see the explanation.

Subtract #2sin(x)# from both sides:

#sqrt(2)tan(x) - 2sin(x) = 0#

Substitute #sin(x)/cos(x)# for #tan(x)#:

#sqrt(2)sin(x)/cos(x) - 2sin(x) = 0#

Multiply both side by #cos(x)#:

#sqrt(2)sin(x) - 2sin(x)cos(x) = 0#

Factor:

#sin(x)(sqrt(2) - 2cos(x)) = 0#

This is true when either factor is zero:

#sin(x) = 0 and cos(x) = sqrt(2)/2#

Use the inverse sine of the first root and the inverse cosine on the second root:

#x = sin^-1(0) and x = cos^-1(sqrt(2)/2)#

#x = {(0 + npi),(pi/4 + 2npi),((7pi)/4 + 2npi):}#

where n is any positive or negative integer, including 0.