Question #aa375

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Answer 1 · 2017-07-05T12:55:15

# x=1/8.#

Recall the Defn. of #arc cos# function :

#arc cos x=theta , |x| <=1 iff costheta=x, theta in [0,pi].#

Let, #theta=arc cos (3/4)., :. costheta=3/4, & 0<=theta<=pi.#

# because costheta=3/4 > 0, :. 0 <=theta<=pi/2.#

Now, given that, #arc cos x=2arc cos (3/4)=2theta.#

# :. cos2theta=x;" but, "cos2theta=2cos^2theta-1=2(3/4)^2-1.#

# :. x=2(3/4)^2-1=1/8.#

Answer 2 · 2017-07-06T04:38:41

cos x = 0.125
arc #x = +- 82^@82#

arccos x = 2 arccos (3/4)
Calculator gives
#cos t = 3/4# --> arc #t = +- 41^@41#
#2t = +- 82^@82#
arc #x = +- 82^@82# --> #cos x = 0.125 = 1/8#