How do you find the exact value of #2sintheta+1=csctheta# in the interval #0<=theta<360#?

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Answer 1 · 2017-01-16T03:24:42

#theta = 30˚, 150˚, 270˚#

Apply #csctheta = 1/sintheta#.

#2sintheta + 1 = 1/sintheta#

#sintheta(2sintheta + 1) = 1#

#2sin^2theta + sin theta - 1 = 0#

Let #t = sintheta#.

#2t^2 + t - 1 = 0#

#2t^2 + 2t - t - 1= 0#

#2t(t + 1) -1(t + 1) = 0#

#(2t -1)(t + 1) = 0#

#t = 1/2 and -1#

#sintheta = 1/2 and sin theta = -1#

#theta = 30˚, 150˚, 270˚#

Hopefully this helps!