Question

Set of maths question for bundle of points ! 

1. Find the number of values of x in the interval [0, 3π] satisfying the equation $2 sin^{2}x + 5sinx - 3 = 0$
2. If $a^{2} sec^{2}A- b^{2} tan^{2}A = c^{2}$, prove that $sinA= \frac{+}{} \sqrt{ \frac{ c^{2}- a^{2} }{ c^{2}- b^{2} } }$
3. Prove that $\frac{1}{cos 290^{o} } + \frac{1}{ \sqrt{3}sin 250^{o} } = \frac{4}{ \sqrt{3} }$

Answer 1

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

Answer 2

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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

°•° proved

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