Trigonometric equation -- real roots

[matrixone]

The number of real roots of the equation

$$2cos \left( \frac {x^2 + x} {6} \right)=2^x + 2^{-x}$$

Answer options are : 0,1,2,∞

My approach :

range of cos function is [-1,1]
thus the RHS of the equation belongs to [-2,2]
So, we have
-2 ≤ 2^x + 2^-x ≤ 2
solving the right inequality, i got 2^x = 1 and that satisfies the left part too
therefore , x can take only one value = 0
Now since this values agrees with the orginal trigonometric equation, we have 1 real root for this equation

So answer is 1

Is my approach/answer correct ?

 

[QuantumQuest]

It is correct. You can also start from the RHS and try to see if $2^{x} + 2^{-x}$ bounds some expression with known value.

 

[fresh_42]

The number of real roots of the equation

$$2cos \left( \frac {x^2 + x} {6} \right)=2^x + 2^{-x}$$

Answer options are : 0,1,2,∞

My approach :

range of cos function is [-1,1]
thus the RHS of the equation belongs to [-2,2]
So, we have
-2 ≤ 2^x + 2^-x ≤ 2
solving the right inequality, i got 2^x = 1 and that satisfies the left part too
therefore , x can take only one value = 0
Now since this values agrees with the orginal trigonometric equation, we have 1 real root for this equation

So answer is 1

Is my approach/answer correct ?

Looks good so far, but how do you get from $2^x +2^{-x}\leq 2$ to $2^x=1$?
I only see $-1 < x < 1$ without pen and paper.

 

[matrixone]

@fresh_42
$2^x +\frac{1}{2^x}\leq 2$
$(2^x)^2 - 2*2^x + 1 \leq 0$
$(2^x - 1)^2 \leq 0$

only the equality condition is feasible for real x

 

[mfb]

Alternatively, you can see that x=0 is a solution to $2^x + 2^{-x} = 2$ and consider the derivatives to show that there are no further solutions. The quadratic equation is more elegant, however.

 

[pasmith]

Looks good so far, but how do you get from $2^x +2^{-x}\leq 2$ to $2^x=1$?
I only see $-1 < x < 1$ without pen and paper.

$2^x + 2^{-x} = 2\cosh(x \ln 2) \geq 2$.