Sec2x= ____. select all that apply

1 answer:

4 0

The best way to solve such questions is to simplify the options and see what they result in:

Simplifying 1st option:
$\frac{1}{ cos^{2}x - sin^{2}x }= \frac{1}{cos(2x)}=sec(2x)
$

Simplifying 2nd option:
$\frac{1}{1- 2sin^{2}x } = \frac{1}{cos(2x)}=sec(2x)$

Simplifying 3rd option:
$\frac{1}{ sin^{2}x - cos^{2}x } = \frac{1}{-cos(2x)} =-sec(2x)$

Simplifying 4th option:
$\frac{1}{2 sin^{2}x }= \frac{ cosec^{2}x }{2}$

So only first two option are the correct ones as they simplify to sec(2x).

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Solve the inequality 2x>30+5/4x

insens350 [35]

Answer:

Step-by-step explanation:

$2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} } \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0$

$x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }$

case~2

$if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0$