Please explain and I also need the answer ASAP... I need help with all of them haha...

Mathematics

1 answer:

Umnica [9.8K]3 years ago

4 0

Answer:

I don't know how to help either I need the same answers as you pls

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Solve the equation on the interval [0,2π). cos^4x=cos^4xcscx

Alisiya [41]

$\bf cos^4(x)=cos^4(x)csc(x)\\\\
-----------------------------\\\\
cos^4(x)=cos^4(x)\cfrac{1}{sin(x)}\implies cos^4(x)=\cfrac{cos^4(x)}{sin(x)}
\\\\\\
cos^4(x)sin(x)=cos^4(x)\implies cos^4(x)sin(x)-cos^4(x)=0
\\\\\\
cos^4(x)[sin(x)-1]=0\to 
\begin{cases}
cos^4(x)=0\to x=cos^{-1}(0)\\
----------\\
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$\bf \measuredangle x = cos^{-1}(0)\implies \measuredangle x = 
\begin{cases}
\frac{\pi }{2}\\\\
\frac{3\pi }{2}
\end{cases}
\\\\\\
\measuredangle x=sin^{-1}(1)\implies \measuredangle x=\frac{\pi }{2}$

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3 years ago

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Step-by-step explanation:

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d1i1m1o1n [39]

Step-by-step explanation:

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HELP !!!! What are the solutions to the quadratic equation 6x^2 + 24x = 0 ( explain )

GREYUIT [131]

$6x(x + 4) = 0$
Now you have 2 case
$6x = 0$
and
$x + 4 = 0$
Solve
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Minus 4 on both sides
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3 years ago