Asked by Luis
solve equation exactly over interval (0, 2ð)
sin(x)= 1-2sin^2(x)
sin(x)= 1-2sin^2(x)
Answers
Answered by
bobpursley
change sin(x) to u.
u=1-2u^2
2u^2+u-1=0
(2u-1)(u+1)=0
u= 1/2 or -1
sin(x)=1/2 or sinx=-1
now solve.
u=1-2u^2
2u^2+u-1=0
(2u-1)(u+1)=0
u= 1/2 or -1
sin(x)=1/2 or sinx=-1
now solve.
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