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Geometry Level 3

What is the sum of all x ∈ [ 0 , π ] x \in [0,\pi ] which satisfy this equation?

sin ⁡ x + 1 2 cos ⁡ x = sin ⁡ 2 ( x + π 4 ) \sin{x}+\frac { 1 }{ 2 } \cos{x}={ \sin }^{ 2 }\left(x+\frac { \pi }{ 4 }\right)

π 6 \frac { \pi }{ 6 } 2 π 1 \frac { 2\pi }{ 1 } 5 π 6 \frac { 5\pi }{ 6 } π 1 \frac { \pi }{ 1 }

Mehul Chaturvedi by Mehul Chaturvedi , 22, India

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3 solutions

Chew-Seong Cheong
Nov 17, 2014

sin ⁡ x + 1 2 cos ⁡ x = sin ⁡ 2 ( x + π 4 ) = 1 2 ( 1 − cos ⁡ ( 2 x + π 2 ) ) = 1 2 ( 1 + sin ⁡ ( 2 x ) ) 2 sin ⁡ x + cos ⁡ x = 1 + 2 sin ⁡ x cos ⁡ x 2 sin ⁡ x cos ⁡ x − 2 sin ⁡ x − cos ⁡ x + 1 = 0 ( 2 sin ⁡ x − 1 ) ( cos ⁡ x − 1 ) = 0 \begin{aligned} \sin x + \frac 12 \cos x & = \sin^2 \left(x + \frac \pi 4 \right) \\ & = \frac 12 \left( 1-\cos \left(2x + \frac \pi 2 \right) \right) \\ & = \frac 12 \left( 1+\sin (2x) \right) \\ 2 \sin x + \cos x & = 1 + 2\sin x \cos x \\ 2\sin x \cos x - 2 \sin x - \cos x + 1 & = 0 \\ (2 \sin x -1)(\cos x - 1) & = 0 \end{aligned}

⟹ { sin ⁡ x = 1 2 ⟹ x = π 6 , 5 π 6 cos ⁡ x = 1 ⟹ x = 0 \implies \begin{cases} \sin x = \frac 12 & \implies x = \frac \pi 6, \ \frac {5\pi}6 \\ \cos x = 1 & \implies x = 0 \end{cases}

Therefore, the required answer is π 6 + 5 π 6 + 0 = π \frac \pi 6 + \frac {5\pi}6 + 0 = \boxed{\pi}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

You use the angle addition formula for cos(2x + pi/2). But pi/2 is such a simple angle to consider you could use the identity cos(theta + pi/2) = -sin(theta) directly by considering the geometry behind the trig functions.

Richard Desper - 4 years, 6 months ago

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Thanks for your comments. I have changed the solution. It was one of my earlier solution.

Chew-Seong Cheong - 4 years, 6 months ago
Sandeep Rathod
Nov 17, 2014

s i n x + 1 2 c o s x = 1 2 ( 1 + 2 s i n x c o s x ) sinx + \frac{1}{2}cosx = \frac{1}{2}(1 + 2sinxcosx)

= c o s x ( 1 2 − s i n x ) = 1 2 − s i n x = cosx(\frac{1}{2} - sinx) = \frac{1}{2} -sinx

( 1 2 − s i n x ) ( c o s x − 1 ) = 0 (\frac{1}{2} - sinx)(cosx - 1) = 0

Thus ,

s i n x = 1 2 , x = π 6 , 5 π 6 sinx = \frac{1}{2} , x = \frac{\pi}{6} , \frac{5\pi}{6}

c o s x = 0 , x = 2 n π cosx = 0 , x = 2n\pi

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Why Level 5 @Mehul Chaturvedi

sandeep Rathod - 6 years, 6 months ago

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exactly ._. i think he is generous enough to gift us 210 points :D

Aritra Jana - 6 years, 6 months ago

(cosx-1)=0 so cosx=1! why 0?

Agastya Chandrakant - 6 years, 5 months ago

You do the hard part, and then solve (cos x-1) = 0 with cos x = 0!!!! But it's OK because you then give the wrong solution for cos x=0. You give the solution for cos x = 1, which is what you really wanted.

Richard Desper - 4 years, 6 months ago
Alessandro Bellia
Oct 28, 2016

sin ⁡ x + 1 2 cos ⁡ x = sin ⁡ 2 ( x + π 4 ) ⇒ 2 sin ⁡ x + cos ⁡ x = 2 sin ⁡ 2 ( x + π 4 ) ⇒ 2 sin ⁡ x + cos ⁡ x = 2 sin ⁡ x cos ⁡ x + 1 ⇒ 4 sin ⁡ 2 x + 4 sin ⁡ x cos ⁡ x + cos ⁡ 2 x = 4 sin ⁡ 2 x cos ⁡ 2 x + 4 sin ⁡ x cos ⁡ x + 1 The two sides are squared ⇒ 4 sin ⁡ 2 x + 1 − sin ⁡ 2 x = 4 sin ⁡ 2 x cos ⁡ 2 x + 1 ⇒ 3 sin ⁡ 2 x = 4 sin ⁡ 2 x cos ⁡ 2 x ⇒ 3 = 4 cos ⁡ 2 x ⇒ cos ⁡ x = ± 3 2 ⇒ x = π 6 o r x = 5 6 π \sin { x } +\frac { 1 }{ 2 } \cos { x } =\sin ^{ 2 }{ \left( x+\frac { \pi }{ 4 } \right) } \\ \\ \Rightarrow 2\sin { x } +\cos { x } =2\sin ^{ 2 }{ \left( x+\frac { \pi }{ 4 } \right) } \\ \\ \Rightarrow 2\sin { x } +\cos { x } =2\sin { x } \cos { x } +1\\ \\ \Rightarrow 4\sin ^{ 2 }{ x } +4\sin { x } \cos { x } +\cos ^{ 2 }{ x } =4\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } +4\sin { x } \cos { x } +1 \text { The two sides are squared} \\ \\ \Rightarrow 4\sin ^{ 2 }{ x } +1-\sin ^{ 2 }{ x } =4\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } +1\\ \\ \Rightarrow 3\sin ^{ 2 }{ x } =4\sin ^{ 2 }{ x } \cos ^{ 2 }{ x } \\ \\ \Rightarrow 3=4\cos ^{ 2 }{ x } \\ \\ \Rightarrow \cos { x } =\pm \frac { \sqrt { 3 } }{ 2 } \\ \\ \Rightarrow x=\frac { \pi }{ 6 } \quad or\quad \quad x=\frac { 5 }{ 6 } \pi

0 Helpful 0 Interesting 0 Brilliant 0 Confused

At a certain point, you divide by sin^2(x) without allowing for possibility that sin x = 0. Thus you neglect the possibility of x = 0. You may have gotten lucky when asked to sum the solutions, since 0 does not contribute to the sum.

Richard Desper - 4 years, 6 months ago

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