Answers

2014-08-09T16:29:04+05:30
1.  2sin²x+5sinx-3 = 0
2sin²x+6sinx-sinx-3 = 0
(sinx+3)(2sinx-1) = 0
sinx = -3 or sinx = 1/2
as -1≤sinx≤1 so sinx = -3 is not the solution 
sinx = 1/2
for x∈[0,3π]
number of solution = 4          
   x = (π/6,5π/6,13π/6,17π/6)

2.  a²sec²x-b²tan²x = c²
a²/cos²x - b²sin²x/cos²x = c²
a² - b²sin²x = c² - c²sin²x
sin²x = (c² - a²)/(c² - b²)
sinx = +-√{(c² - a²)/c² - b²)}
3. 1/cos290 + 1/√3sin250
  (√3sin250 + cos290)/√3sin250cos290
{cos(270+20) + √3sin(270-20)}/{√3/2(sin(290+250) + sin(250-290))}
2{sin20/2 - √3cos20/2}/{-√3/2sin 40}
2{sin(20-60)}/(-√3/2 sin40)
2/{√3/2}
4/√3  prove
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