3 years ago

USIU

Mathematics

2 answers:

topjm [15]3 years ago

8 0

Answer:

Test Cross , 50% to 50%

Step-by-step explanation:

just took the test yw

Slav-nsk [51]3 years ago

5 0

Answer:

test cross and 50 50

light work

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Cot^4θ-4cot^2θ-5=0 <br>solve for θ

stich3 [128]

$\cot^4\theta-4\cot^2\theta-5=(\cot^2\theta-5)(\cot^2\theta+1)=0$

$\cot^2\theta-5=0\implies\cot^2\theta=5\implies\cot\theta=\pm\sqrt5$

Recall that $\cot\theta=\cot(\theta+\pi)$ , which is to say $\cot x$ has period $\pi$ . This in turn means that $\cot\theta=0$ will have the same solutions as $\cot(\theta+n\pi)=0$ for any integer $n$ . So the general solution to the first case is

$\cot\theta=\pm\sqrt5\implies\theta=\cot^{-1}(\pm\sqrt5)+n\pi$

where $n$ is any integer.

On the other hand,

$\cot^2\theta+1=0\implies\cot^2\theta=-1$

but $x^2\ge0$ for any value of $x$ , so this equation has no (real) solutions for $\theta$ .

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3 years ago