Mathematics: Find all solutions in the interval [0,2pi). sec^2x-2=tan^2x

Find all solutions in the interval [0,2pi). sec^2x-2=tan^2x

Ответ:

gonzalesnik

16.01.2022

$\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad \qquad sec(\theta)=\cfrac{1}{cos(\theta)}
\\\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-----------------------------\\\\
sec^2(x)-2=tan^2(x)\qquad [0,2\pi )
\\\\\\
\cfrac{1^2}{cos^2(x)}-2=\cfrac{sin^2(x)}{cos^2(x)}\implies \cfrac{1}{cos^2(x)}-2-\cfrac{sin^2(x)}{cos^2(x)}=0$

$\bf \cfrac{1-2cos^2(x)-sin^2(x)}{cos^2(x)}=0\implies 1-2\underline{cos^2(x)}-sin^2(x)
\\\\\\
1-2\underline{[1-sin^2(x)]}-sin^2(x)=0\implies 1-2+2sin^2(x)-sin^2(x)=0
\\\\\\
-1+sin^2(x)=0\implies sin^2(x)=1\implies sin(x)=\pm\sqrt{1}
\\\\\\
sin(x)=\pm 1\implies \measuredangle x=sin^{-1}(\pm1)\implies \measuredangle x=
\begin{cases}
\frac{\pi }{2}\\\\
\frac{3\pi }{2}
\end{cases}$

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Ответ:

rosa1423

21.03.2023

Mathematics

General Equation of a circle with centre (a, b) and radius r

(x − a)² + (y − b)² = r²

Substitute (6,0) and r = 6

(x - 6)² + (y - 0)² = 6²
(x - 6)² + y² = 36

(x - 6)² + y² = 36 (Answer A)

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Find all solutions in the interval [0,2pi). sec^2x-2=tan^2x

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