Solving \(\displaystyle\sqrt{{{2}}}={\frac{{{\cos{{\left({x}\right)}}}}}{{{\cos{{\left({2}{x}\right)}}}}}}\) for \(\displaystyle{0}\leq{\left\lbrace{x}\right\rbrace}\leq{\left\lbrace{2}\pi\right\rbrace}\)? I can't find

Hanzikoval1pa

Hanzikoval1pa

Answered question

2022-04-03

Solving 2=cos(x)cos(2x)
for 0{x}{2π}? I can't find any trigonometric identity to solve it.

Answer & Explanation

mistemePietsffi

mistemePietsffi

Beginner2022-04-04Added 8 answers

We have cosine of sum:

Letting A=B=x gives the double angle formula for cosine:

Using the fact that sin2(x)+cos2(x)=1, write sin2(x)=1cos2(x) to give:
cos(2x)=cos2(x)[1cos2(X)]=2cos2(x)1
Then, your equation becomes:
2=cos(x)2cos2(x)1
Rearranging gives:
22cos2(x)cos(x)2=0
Let y=cos(x). Then:
22y2y2=0
which is a quadratic equation that you hopefully can solve.

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