A geometry problem by Ahmad Sazidy

Geometry Level 2

If $1 - { e }^{ 2x+y }=0$ and $\sin x = \sin y$ , Then which of these pairs satisfy the value of $x$ and $y$ respectively?

Angles are measured in radian.

x = \pi, y =-2\pi

x=1 ,y=-2

x=180 ,y=-360

x=90, y=-180

1 solution

Ahmad Sazidy
Aug 21, 2015

Here,

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

${e}^{2x+y}$ =1

ln( ${e}^{2x+y}$ )=ln(1)

2 ${x}+y$ =0

${y}$ =-2 ${x}$

Now from the second equation,

$sin{x}=sin{y}$

$sin{x}=sin(-2x)$

$sin{x}=-2sin{x}cos{x}$

$sin{x}(1+2cos{x})=0$

So one solution is $sin{x}=0$

so, $x= 0, \pi , .....$

we take ${x}=\pi$ so $y=-2{x} =-2\pi$

Where does it say that angles are taken in radians?

Minjae Son - 5 years, 9 months ago

you have to take the value of x in radian. you can not put x=180 degree in first equation.

Ahmad Sazidy - 5 years, 9 months ago

Yeah... You still need to specify that they are radians. Somemore $\pi \ne 3.1416$ .

Pi Han Goh - 5 years, 9 months ago

@Pi Han Goh – In trigonometric and algebraic equations x,y etc. variables are specified in radian by convention. if it is in degree then specified as sin(x°) . However it is specified now. and the values were rounded not exact. thank you

Ahmad Sazidy - 5 years, 9 months ago

A geometry problem by Ahmad Sazidy

1 solution

1- e 2 x + y {e}^{2x+y} e 2 x + y =0

e 2 x + y {e}^{2x+y} e 2 x + y =1

ln( e 2 x + y {e}^{2x+y} e 2 x + y )=ln(1)

2 x + y {x}+y x + y =0

y {y} y =-2 x {x} x

Now from the second equation,

s i n x = s i n y sin{x}=sin{y} s i n x = s i n y

s i n x = s i n ( − 2 x ) sin{x}=sin(-2x) s i n x = s i n ( − 2 x )

s i n x = − 2 s i n x c o s x sin{x}=-2sin{x}cos{x} s i n x = − 2 s i n x c o s x

s i n x ( 1 + 2 c o s x ) = 0 sin{x}(1+2cos{x})=0 s i n x ( 1 + 2 c o s x ) = 0

So one solution is s i n x = 0 sin{x}=0 s i n x = 0

so, x = 0 , π , . . . . . x= 0, \pi , ..... x = 0 , π , . . . . .

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