How do you solve $sec ((5 theta)/4) =2$ from 0 to 2pi?

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Answer 1 · 2016-10-11T16:43:56

$S={48^@,240^@,336^@}$

$sec ((5theta)/4)=2$

$1/cos ((5theta)/4)=2$

$1=2 cos((5theta)/4)$

$1/2 = cos((5theta)/4)$

$cos^-1 (1/2) =((5theta)/4)$

$+- 60^@ + 360n^@=(5theta)/4$

$4/5(+- 60^@ + 360n^@)=theta$

$+-48^@+288n^@=theta$

$n=0, theta=+- 48^@$

$n=1, theta = 336^@, 240^@$

$S={48^@,240^@,336^@}$

How do you solve sec ((5 theta)/4) =2sec ((5 theta)/4) =2 from 0 to 2pi?