Solve this equation for x \in (0,\frac{\pi}{2}) \cos(2x)\cos(x-\frac{\pi}{6})=\sin(2x)\sin(\frac{\pi}{6}-x) I gave a try

Victor Wall 2022-01-15 Answered
Solve this equation for x(0,π2)
cos(2x)cos(xπ6)=sin(2x)sin(π6x)
I gave a try using the cos(ab) and sin(ab) formulas, but it seems the problem complicated a little bit more.
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Answers (3)

temzej9
Answered 2022-01-17 Author has 30 answers
cos(2x)cos(π6x)sin(2x)sin(π6x)=0
cos(2x+π6x)=0
(using cos(a+b)=cosacosbsinasinb)
x+π6=π2
x=π3
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RizerMix
Answered 2022-01-20 Author has 438 answers
It is not complicated, the equation become cos(2x)cos(xπ6)+sin(2x)sin(xπ6)=0 that is equivalent to cos(x+π6)=0 Thus the solution is x+π6=π2+kπ x=π3+kπ
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user_27qwe
Answered 2022-01-23 Author has 208 answers
The equation is cos(2x)cos(xπ6)+sin(2x)sin(xπ6=0. This is same as cos(2x(xπ6))=0 or cos(x+π6)=0. So x+π6=(2n+1)π2 for some integer n. For x(0,π2) we must have n=0 so x=π3
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