How do you solve #2sin^2xtanx=tanx#?

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Answer 1 · 2016-09-22T12:12:06

#x = 0, (pi) / (4)#

We have: #2 sin^(2)(x) tan(x) = tan(x)#

#=> 2 sin^(2)(x) tan(x) - tan(x) = 0#

#=> tan(x) (2 sin^(2)(x) - 1) = 0#

#=> tan(x) = 0#

#=> x = arctan(0)#

#=> x = 0#

or

#=> 2 sin^(2)(x) - 1 = 0#

#=> sin^(2)(x) = (1) / (2)#

#=> sin(x) = (sqrt(2)) / (2)#

#=> x = arcsin((sqrt(2)) / (2))#

#=> x = (pi) / (4)#

Therefore, the solutions to the equation are #x = 0# and #x = (pi) / (4)#.