How do you solve #sinx+4cscx+5=0# and find all solutions in the interval #0<=x<360#?

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Answer 1 · 2016-10-03T02:04:58

#x = 270°#

Given:
#sin(x) + 4csc(x) + 5 = 0; 0° <= x < 360°#

Multiply both sides by sin(x):

#sin²(x) + 5sin(x) + 4 = 0; 0° <= x < 360°#

Factor:

#(sin(x) + 4)(sin(x) + 1) = 0; 0° <= x < 360°#

#sin(x) = - 4# and #sin(x) = -1; 0° <= x < 360°#

We must discard the #sin(x) = -4# answer, because the sine function only returns values between -1 and 1 inclusive.

That leaves us with:

#sin(x) = -1; 0° <= x < 360°#

Take the inverse sine of both sides:

#x = sin^-1(-1); 0° <= x < 360°#

#x = 270°#