Mathematics.Geometry & Trigonometry

Mathematics Forum

Author	Message / Information
BoySetsFire Quote \| Reply \|	Trigonometric Equation posted on: 11/9/2006 9:31:01 AM Hi! Can anyone help me or show me a way to solve the following trigonometric equation? ( sin 3x + sqrt3 cos 3x)^2 - 2 cos 14x = 2
herbert Quote \| Reply \|	Trigonometric Equation replied on: 11/16/2006 4:09:43 AM solving the squared bracket using the relations sin(x)^2 + cos(x)^2 =1 and 2sin(3x)cos(3x) = sin(6x) and cos(3x)^2 =(1+cos(6x))/2 you will arrive at something like acos(6x)+bsin(6x)=2*cos(14x) a, b beeing constants after reducing the expression to its minumum form I strongly assume that there is a mistake in the problem posed as cos(14x) cannot be resolved in parts containing only multiples of 6 therefore it is your turn to solve the problem by trial and error or with the help of a good calculator
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