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3 years ago

Find the number of real number solutions for the equation. x^2+0x-1=0

Mathematics

1 answer:

Kazeer [188]3 years ago

3 0

Answer:

-1

Step-by-step explanation:

Given equation :

$x^{2} +0x-1$ = 0

$x^{2} -1 = 0\\x^{2} =1\\x=\sqrt{1}$

x= ±1

x= 1 and x= -1

±1 lies in solution set of real numbers !

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Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=9%5C%3A%20%20%5C%3A%20%20%5Cfrac%7B%20%5Csin%28%20%5Ctheta%29%20%7D%7B1%20%2B%20%20%5Ccos%28%2

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Option (b) is your correct answer.

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given Trigonometric expression is

$\rm :\longmapsto\:\dfrac{sin\theta }{1 + cos\theta }$

So, on rationalizing the denominator, we get

$\rm \: = \: \dfrac{sin\theta }{1 + cos\theta } \times \dfrac{1 - cos\theta }{1 - cos\theta }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}}$

So, using this, we get

$\rm \: = \: \dfrac{sin\theta (1 - cos\theta )}{1 - {cos}^{2}\theta }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this identity, we get

$\rm \: = \: \dfrac{sin\theta (1 - cos\theta )}{{sin}^{2}\theta }$

$\rm \: = \: \dfrac{1 - cos\theta }{sin\theta }$

<u>Hence, </u>

$\\ \red{\rm\implies \:\boxed{\tt{ \rm \:\dfrac{sin\theta }{1 + cos\theta } = \: \dfrac{1 - cos\theta }{sin\theta } }}} \\$

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