Solve the multiple-angle equation in the interval [0, 2π): sec2x = 2

\[ \cos(2 x) =\frac 1 {\sec (2x)}=\frac 1 2 \] Finish it from here

are u talking to the asker?

Yes

ok

What angle has cosine 1/2 in your given interval?

My possible answers are A. x = π/8, 7π/8, 9π/8, 15π/8 B. x = π/12, 5π/12, 13π/12, 17π/12 C. x = π/6, 5π/6, 7π/6, 11π/6 D. x = π/4, 3π/4, 5π/4, 7π/4 E. x = π/3, 2π/3, 4π/3, 5π/3

\[ \cos(2 x)=\frac 1 2\\ 2 x = \pm \frac \pi 3 + 2 k \pi\\ x = \pm \frac \pi 6 + k \pi\\ \] Which answer to choose from your choices?

Yeah. How do you get from what you just wrote to one of my options?

vary k over \[ k=0,\, \pm 1, \, \pm2, \cdots \]

you can take \[ \frac \pi 6 + 0=\frac \pi 6\\ \frac \pi 6 +1 \pi=\frac {7\pi} 6\\ -\frac \pi 6 +1 \pi=\frac {5\pi} 6\\ -\frac \pi 6 +2 \pi=\frac {11\pi} 6\\ \]

Did you understand it?

Oooohh. Okay. Thanks for explaining. That was very helpful

yw

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