Are you ready for JEE-3?

Geometry Level 5

The number of all possible values of θ \theta , where 0 < θ < π 0<\theta<\pi , for which the system of equations: ( y + z ) cos ⁡ 3 θ = ( x y z ) sin ⁡ 3 θ x sin ⁡ 3 θ = 2 cos ⁡ 3 θ y + 2 sin ⁡ 3 θ z ( x y z ) sin ⁡ 3 θ = ( y + 2 z ) cos ⁡ 3 θ + y sin ⁡ 3 θ (y+z)\cos 3\theta=(xyz)\sin 3\theta \\ x\sin 3\theta=\dfrac{2\cos 3\theta}{y}+\dfrac{2\sin 3\theta }{z}\\ (xyz)\sin 3\theta=(y+2z)\cos 3\theta +y\sin 3\theta have a solution ( x 0 , y 0 , z 0 ) (x_0,y_0,z_0) with y 0 z 0 ≠ 0 y_0z_0\neq0 , is:

8 0 3 2 5

Rishabh Jain by Rishabh Jain , 22, Germany

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rishabh Jain
Apr 10, 2016

This question appeared in JEE 2010 paper

Since y 0 z 0 ≠ 0 ⟹ y 0 , z 0 ≠ 0 y_0z_0\neq 0 \implies y_0,z_0\neq 0 . Therefore we can write equations as:

x sin ⁡ 3 θ − cos ⁡ 3 θ y − cos ⁡ 3 θ z = 0... ( 1 ) x sin ⁡ 3 θ − 2 cos ⁡ 3 θ y − 2 sin ⁡ 3 θ z = 0.... ( 2 ) x sin ⁡ 3 θ − 2 cos ⁡ 3 θ y − cos ⁡ 3 θ + sin ⁡ 3 θ z . . . ( 3 ) x\sin 3\theta -\dfrac{\cos 3\theta}{y}-\dfrac{\cos 3\theta}{z}=0...(1)\\ x\sin 3\theta-\dfrac{2\cos 3\theta}{y}-\dfrac{2\sin 3\theta }{z}=0....(2)\\ x\sin 3\theta-\dfrac{2\cos 3\theta}{y}-\dfrac{\cos 3\theta+\sin 3\theta}{z}...(3)

Subtracting ( 2 ) (2) and ( 3 ) (3) we get:-

2 sin ⁡ 3 θ = cos ⁡ 3 θ + sin ⁡ 3 θ 2\sin 3\theta=\cos 3\theta +\sin 3\theta ⟹ sin ⁡ 3 θ = cos ⁡ 3 θ \implies \sin 3\theta=\cos 3\theta ⟹ tan ⁡ 3 θ = 1 \implies \tan 3\theta=1 which obviously has 3 \boxed 3 solutions in ( 0 , π ) (0,\pi) namely π 12 , 5 π 12 , 9 π 12 \dfrac{\pi}{12},\dfrac{5\pi}{12},\dfrac{9\pi}{12} .

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Why is this level 5? LOL

Nihar Mahajan - 5 years, 2 months ago

Log in to reply

I set the level 3 for the problem but it will surely take time for the problem to gain its true level..

Rishabh Jain - 5 years, 2 months ago

Something I don't understand. The problem asks for values of theta that yield solutions to the system of equations - but you have not shown that the system has solutions at the possible values of theta. In particular, because cos and sin are equal here, we can divide all equations by the trig functions. Then the last and first equations become xyz = y + z and xyz = 2y + 2z. These are inconsistent unless y, z = 0. Can you explain where my reasoning is wrong here?

Jacob Swenberg - 4 years, 8 months ago

Log in to reply

No, you could have x = 0 , y = − z , x = 0, y = -z, which is in fact what is implied by the solution above. I agree that the solution is not complete as written, though.

Patrick Corn - 3 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...