How do you solve #2sinx -1 = cscx#?
1 Answer
Jan 12, 2018
Explanation:
Rewrite in terms of sine.
#2sinx - 1 = 1/sinx#
#sinx(2sinx - 1) = 1#
#2sin^2x - sinx - 1 = 0#
#2sin^2x - 2sinx + sinx - 1 = 0#
#2sinx(sinx - 1) + 1(sinx- 1) = 0#
#(2sinx + 1)(sinx - 1) = 0#
#sinx = -1/2 or sinx = 1#
When
When
Hopefully this helps!