How do you solve sec $x$ $x$ = cosec $\frac{3}{4} π$ $3/4pi$ for the range value of $0 \leq x \leq 2 π$ $0 <= x <= 2pi$ ?

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How do you solve sec $x$ $x$ = cosec $\frac{3}{4} π$ $3/4pi$ for the range value of $0 \leq x \leq 2 π$ $0 <= x <= 2pi$ ?

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Answer 1 · 2018-04-01T01:59:07

$x = \frac{π}{4}, \frac{7 π}{4}$ $x=pi/4, (7pi)/4$

Explanation:

First, let's evaluate $csc (\frac{3 π}{4}) .$ $csc((3pi)/4).$ Knowing that $csc x = \frac{1}{sin x}, csc (\frac{3 π}{4}) = \frac{1}{sin (\frac{3 π}{4})}$ $cscx=1/sinx, csc((3pi)/4)=1/sin((3pi)/4)$

Now, $sin (\frac{3 π}{4}) = \frac{\sqrt{2}}{2}, csc (\frac{3 π}{4}) = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}}$ $sin((3pi)/4)=sqrt(2)/2, csc((3pi)/4)=1/(sqrt(2)/2)=2/sqrt(2)$

So, we are really just being asked to solve

$sec x = \frac{2}{\sqrt{2}}$ $secx=2/sqrt(2)$

Recalling that $sec x = \frac{1}{cos x} :$ $secx=1/cosx:$

$\frac{1}{cos x} = \frac{2}{\sqrt{2}}$ $1/cosx=2/sqrt(2)$

Invert both sides to get the cosine on top:

$cos x = \frac{\sqrt{2}}{2}$ $cosx=sqrt(2)/2$

In the interval $[0, 2 π],$ $[0, 2pi],$ this holds true for $x = \frac{π}{4}, \frac{7 π}{4}$ $x=pi/4, (7pi)/4$

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How do you solve sec $x$ $x$ = cosec $\frac{3}{4} π$ $3/4pi$ for the range value of $0 \leq x \leq 2 π$ $0 <= x <= 2pi$ ?

1 Answer

Explanation:

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How do you solve sec xxx = cosec 3/4pi34π3/4pi for the range value of 0 <= x <= 2pi0≤x≤2π0 <= x <= 2pi ?

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Explanation:

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How do you solve sec $x$ $x$ = cosec $\frac{3}{4} π$ $3/4pi$ for the range value of $0 \leq x \leq 2 π$ $0 <= x <= 2pi$ ?