Question

For which values of does the following equation have exactly one solution?

Answer 1

Answer:

Consider the equation

x2+2xcost√+1sint=22–√

Observe that all solutions must be in Quadrant I or Quadrant IV in order for the cosine to be positive.

We have…

x2+2xcost√+1sint−22–√=0

This is the standard form of a quadratic equation with

a=1

b=2cost√

c=1sint−22–√

It has a single (double) solution whenever its discriminant is zero.

Δ=b2−4ac=4cost−4(1)(1sint−22–√)

=4cost−4sint+82–√

=4sint−4cost+8sintcost2–√sintcost

This expression is zero when its numerator is zero.

4sint−4cost+8sintcost2–√=0

sint−cost+2sintcost2–√=0

2sintcost2–√=cost−sint

sin(2t)2–√=cost−sint

2sin2(2t)=1−2sintcost

2sin2(2t)=1−sin(2t)

2sin2(2t)+sin(2t)−1=0

sin(2t)=12 , since sin(2t)=−1 yields a value of t in the second quadrant, for which there is no solution.

2t=π6+2nπ

t=π12+nπ

Check:

If n is odd, such as when t=13π12 , cost < 0, and its square root is not real. However, when n is even, the result follows.

The solution is

t=π12+2nπ

Step-by-step explanation:

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Answer 2

Given :   $x^2 +\frac{1}{ \sqrt{\cos t} } 2x+\frac{1}{\sin t} =2\sqrt{2}$

t ∈ R

To Find : Values of t for which equation have exactly one solution

Solution:

Quadratic equation is of the form ax²+bx+c=0  where a  , b and c are real also  a≠0.

D =  b²-4ac is called discriminant.

D >0 roots are real and distinct

D =0 roots are real and equal

D < 0 roots are imaginary ( not real ) and different

$x^2 +\frac{1}{ \sqrt{\cos t} } 2x+\frac{1}{\sin t} =2\sqrt{2}$

$\implies x^2 +\frac{1}{ \sqrt{\cos t} } 2x+\frac{1}{\sin t} -2\sqrt{2}=0$

a = 1 , b  = 2/√cost   , c = (1/sin t )  - 2√2

D = 0

( 2/√cost )² - 4(1)(1/sin t  - 2√2) = 0

=> 4/cost  - 4/sin t  + 8√2 = 0

=> 1/cost  - 1/sin t  + 2√2 = 0

=> sint - cost  + 2√2costsin t =  0

=>  cost  - sint  = 2√2costsint

=> (1/√2) cost  - (1/√2)sint = 2costsint

Sin(A - B) = sinAcosB - cosAsinB

=> Sin(π/4 - t) = sin2t

π/4 - t = nπ + (-1)ⁿ2t

n = 2k + 1

π/4 - t = (2k + 1)π  - 2t

=> t = (2k + 1)π - π/4

n = 2k

=> π/4 - t = 2kπ +  2t

=> 3t = π/4 - 2kπ

=> t = π/12 - 2kπ/3

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