The general solution of the equation sec⁡4θ−sec⁡2θ=2 is θ=

The general solution of the equation sec4θsec2θ=2 is θ=

  1. A

    (2n+1)π2

  2. B

    (2n+1)π5

  3. C

    (2n+1)π3

  4. D

    (2n-1)π10

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    Solution:

    1cos4θ1cos2θ=2

        cos2θcos4θ=2cos4θcos2θ=cos6θ+cos2θ    cos6θ+cos4θ=0    2cos5θcosθ=0cosθ=0θ=(2n+1)π2,cos5θ=0θ=(2n+1)π10

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