\(\displaystyle{{\sin}^{{2}}{x}}{{\cos}^{{2}}{x}}+{\sin{{x}}}{\cos{{x}}}-{1}={0}\)

ab0utfallingm1z2

ab0utfallingm1z2

Answered question

2022-03-30

sin2xcos2x+sinxcosx1=0

Answer & Explanation

Kailee Castro

Kailee Castro

Beginner2022-03-31Added 8 answers

Hint:
Observe the roots of t2+2t4=0 are 1±5 (no denominator), and their absolute values are greater than 1.
Dixie Reed

Dixie Reed

Beginner2022-04-01Added 15 answers

Another way to continue this problem lies in realizing that since 5±1>1 , we must have that x
You can begin looking for complex solutions to this problem by utilizing the fact that
sin2x=e2ixe2ix2i
and solving yet another quadratic equation in terms of a complex exponential.Applying the complex logarithm
logz=log|z|+iArg z
will give you your solutions.

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