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Trigonometry equalities, inequalities and expressions - sin, cos, tan, cot
by Guest » Wed Jun 23, 2021 10:06 am
I don’t understand how to solve it and in what sequence?
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Guest
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by Guest » Tue Aug 31, 2021 11:27 am
sin(2x)= 2 sin(x)cos(x) and cos(2x)= cos^2(x)- sin^2(x)
so this equation is
$2\sqrt{3}sin(x)cos(x)- cos^2(x)+ sin^2(x)+ 2\sqrt{3} sin(x)+ 2 cos(x)= a$
We can write that as
$sin^2(x)+ 2\sqrt{3}(cos(x)+ 1)sin(x)+ (2cos(x)- cos^2(x)- a)= 0$
Use the quadrtic formula- the solutions to $ax^2+ bx+ c= 0$ are $x= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$-
with $a= 1$, $b= 2\sqrt{3}(cos(x)+ 1)$, and $c= 2cos(x)- cos^2(x)- a$ to get an equation relating sin(x) and cos(x) and solve that equation.
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Guest
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by Guest » Wed Sep 01, 2021 6:02 am
Thanks!
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