How do you solve #10cos(x)^2-3sin(x)+9=0#?

1 Answer
Mrigansh M. · Becca M.
Sep 11, 2015

#sinx= 0.2#
#sinx= -0.5#

Explanation:

Well according to me the equation given by you has a slight mistake the real equation is #-10cos^2x - 3sinx + 9 = 0#, notice the negative sign in front of 10.

Now this equation can be solved by converting #cos^2x# into #(1-sin^2x)# and proceed as follows:

#-10(1-sin^2x)+3sinx+9=0#
#-10-10sin^2x+3sinx+9=0#
#-10sin^2x+3sinx-1=0#
#10sin^2x-3sinx-1=0#

Now you can solve for #sinx# by middle term splitting—you would get two values:

  1. #sinx = 1/2#

  2. #sinx=-1/5#

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