How do you find all solutions of #sin(x/2)+cosx-1=0# in the interval #[0,2pi)#?

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Answer 1 · 2017-07-14T16:32:47

#x=0, pi/3, 5pi/3.#

#sin(x/2)+cosx-1=0.#

# :. sin(x/2)=1-cosx=2sin^2(x/2).#

# :. sin(x/2)-2sin^2(x/2)=0.#

# :. sin(x/2){1-2sin(x/2)}=0.#

# :. sin(x/2)=0, or, sin(x/2)=1/2=sin(pi/6).#

#"But, "x in [0,2pi) :. 0lexlt2pi :. 0lex/2ltpi.#

# :. sin(x/2)=0 rArr x/2=0 :. x=0.#

Similarly, #sin(x/2)=sin(pi/6) rArr x/2=pi/6, pi-pi/6=5pi/6.#

#:. x=pi/3, 5pi/3.#

Altogether, #x=0, pi/3, 5pi/3.#