Question #f3dd9

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Question #f3dd9

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Answer 1 · 2016-07-18T04:37:05

$63^{\circ} 43, and 243^{\circ} 43$ $63^@43, and 243^@43$

Explanation:

tan x - sec x = 1
$\frac{sin x}{cos x} - \frac{1}{cos x} = 1$ $(sin x)/(cos x) - 1/(cos x) = 1$
$\frac{sin x - cos x}{cos x} = 1$ $(sin x - cos x)/(cos x) = 1$
sin x - cos x = cos x
sin x - 2cos x = 0
Call t the arc whose tan t = 2 --> $t = 63^{\circ} 43$ $t = 63^@43$
$sin x - (\frac{sin t}{cos t}) cos x = 0$ $sin x - ((sin t)/(cos t))cos x = 0$
sin x.cos t - sin t.cos x = sin (x - t)= 0
sin (x - 63.43) = 0
There are 2 solutions for the arc $(x - 63^{\circ} 43)$ $(x - 63^@43)$
a. x - 63.43 = 0 --> x = 63^@43
b. x - 63.43 = 180 --> x = 180 + 63.43 = 243^@43
Answers for $(0, 2 π) :$ $(0, 2pi):$
$63^{\circ} 43$ $63^@43$ , and $243^{\circ} 43$ $243^@43$

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Question #f3dd9

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