Equations and Graphs of Trigonometric Functions. Determine the solutions to the equation 4sin2(x+45)=3 over the interval 0<x<360 Algebreically?

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Answer 1 · 2018-07-27T13:17:36

#x ~~ 20.7^@ and x ~~ 339.3^@#

We have the equation

#4sin 2(x+45^@) = 3#
#4 sin(2x+90^@) = 3#

Remember the sum identity for sine:

#sin(a+b) = sinacosb+cosasinb#

#=> sin(2x+90^@) = sin2xcos90^@+cos2xsin90^@=#

#=sin2x * 0 + cos2x * 1 = cos2x#

#:. 4 cos2x = 3=> cos2x = 3/4#

Now, there is no nice form of the answer. We might as well accept that

#2x = arccos(3/4)#

However, as the inverse cosine goes from #0^@# to #180^@#, these are not all the solutions; A more general solution would be

#2x=arccos(3/4)+360^@n#, for #n in ZZ#.

Even then, since the cosine function is even, meaning that

#cos(-2x) = cos(2x) = 3/4#

We reach the additional solutions:

#2x = +- arccos(3/4) + 360^@n#

Out of the complete set of solutions, only two are located in the interval #(0^@, 360^@)#:

#x = arccos(3/4) and x = -arccos(3/4) + 360^@#

#arccos(3/4) ~~ 20.7^@ => -arccos(3/4) + 360^@ = 339.3^@#