Solve #sin^2(x)=cos(x# )for all solutions 0≤x<2π. Ask for X? Give your answers accurate to 2 decimal places, as a list separated by commas

1 Answer
Nghi N.
Jun 20, 2018

#51^@83, 308^@17#

Explanation:

#sin^2 x = cos x#
#(1 - cos^2 x) = cos x#
#cos^2 x + cos x - 1 = 0#.
Solve this quadratic equation for cos x.
#D = d^2 = b^2 - 4ac = 1 + 4 = 5# --> #d = +- sqrt5#
There are 2 real roots:
#cos x = -b/(2a) +- d/(2a) = - 1/2 +- sqrt5/2 = (-1 +- sqrt5)/2#
a. #cos x = (-1 - sqrt5)/2 = -3.24/2 = -1.62# (rejected as < - 1)
b. #cos x = (-1 + sqrt5)/2 = 0.62#
Calculator and unit circle give 2 solutions for x -->
#x = +- 51^@83#
Note. x = - 51.83 is co-terminal to #x = 360 - 51.83 = 308^@17#
Answers for #(0, 2pi)#
#51^@83, 308^@17#
Check by calculator.
x = 308.17 --> #sin^2 x = 0.62# --> cos x = 0.62. Proved

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