Question

What is the solution to the equation `(cos x)/(sin2x) = 5/7` for `0<=x<=4pi`?

Answer 1

#44^@43; 90^@; 155^@57; 270^@; 224^@43; 450^@; 515^@57; 630^@#

Explanation:

Cross multiply:
5sin 2x = 7cos x
5sin 2x - 7cos x = 0
Use trig identity: sin 2x = 2sin x.cos x.
10sin x.cos x - 7cos x = 0
cos x (10 sin x - 7) = 0
a/ cos x = 0 --> Unit circle gives two answers:
`x = pi/2 + 2kpi`, and `x = (3pi)/2 + 2kpi`
b. 10 sin x = 7 --> `sin x = 7/10` --> Calculator and unit circle
give 2 answers:
`x = 44^@43 + k360^@`, and `x = 155^@57 + k360^@`
For answers in (360, 720) add `360^@` (k = 1)
Answers for `(0, 720^@)`
#44^@43, 90^@, 155^@57, 270^@, 224^@43, 450^@, 515^@57, 630^@#