Question

If 5 sin 0+12 cos 0 =
26(a-3)/
(4-3a)
has a real solution for 0, then complete 'a' is​

Answer 1

: If cos = , then will lie in 1st or 4th quadrant. ... Solution: 3 – 2cos q – 4 sin q – cos 2q + sin 2q = 0 ... Þ 3 tan 2q +12cot2q has minimum value 12.

Answer 2

Answer:

12/26 = a-3/4-3a

6(4-3a) = 13(a-3)

24-18a=13a-39

63=31a

a=63/31

Cheers!