How do you solve $2cos^2x = sinx + 1$ from $[0,2pi]$ ?

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Answer 1 · 2015-08-18T15:54:30

x= $pi/6, (5pi)/6$ , $(3pi)/2$

Re write it as $2(1- sin^2x)= sinx +1$ ,

$2sin^2x + sinx -1=0$

(2sinx -1)(sinx +1)=0

sin x= $1/2$ , -1

sin x = $1/2$ would have x= $pi/6, (5pi)/6$ in the prescribed interval

sin x= -1 would have x= $(3pi)/2$

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