How do you solve #2sin^2x-5sinx-3 = 0#?

1 Answer
Nghi N
May 22, 2018

#x = (7pi)/6 + 2kpi#
#x = (11pi)/6 + 2kpi#

Explanation:

#f(x) = 2sin^2 x - 5sin x - 3 = 0#
Solve this quadratic equation for sin x.
#D = d^2 = b^2 - 4ac = 25 + 24 = 49# --> #d = +- 7#
There are 2 real roots:
#sin x = -b/(2a) +- d/(2a) = 5/4 +- 7/4# -->
#sin x = - 2/4 = - 1/2#, and sin x = 3 (rejected as > 1)
#sin x = - 1/2#
Trig table and unit circle give 2 solutions for x:
#x = - pi/6 + 2kpi#, or #x = (11pi)/6 + 2kpi# (co-terminal)
and #x = pi - (-pi/6) = (7pi)/6 + 2kpi#

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
27849 views around the world
You can reuse this answer
Creative Commons License