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Could you please give me some hint how to solve this trigonometric equation:

$$ x\left(\sin x+\cos x\right)=1$$

Since $\sin x+ \cos x= \sin x+ \sin\left(\frac {\pi} 2-x\right)=2\sin\frac{\pi}4cos\left(x-\frac{\pi}4\right)=\sqrt2\cos\left(x-\frac{\pi}4\right)$

The equation transformed to $\sqrt2x\cos\left(x-\frac{\pi}4\right)=1$.

And I have no idea how to proceed.

The source of the problem is to find out if Wronskian of $sinx-cosx,sinx-x$ is equal to zero for any $x \in (0,2\pi)$.

Thanks.

user97484
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    I doubt there is a closed form solution of this problem. Numeric solutions are your best shot. – 5xum Oct 01 '14 at 09:21
  • In addition, take a look to the graph of $F(x) = x (\sin{x} + \cos{x} ) - 1$ and seek for points where $F = 0$: (link: http://www.wolframalpha.com/input/?i=plot+x*%28sin%28x%29+%2B+cos%28x%29+%29+-+1) – Dmoreno Oct 01 '14 at 09:24
  • I think you can only solve it numerically. – UserX Oct 01 '14 at 09:32
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    @user97484, Can you please disclose the source of the problem as this does not seem to be a typical Trigonometric Problem. – lab bhattacharjee Oct 01 '14 at 09:42
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    I wanted to find out if Wronskian of $sinx-cosx,sinx-x$ is equal to zero for any $x \in (0,2\pi)$. – user97484 Oct 01 '14 at 09:53
  • You know that we cannot solve analytically $x=\cos(x)$. The only solution will be Newton method. If you want me to elaborate,just post. – Claude Leibovici Oct 01 '14 at 10:22

1 Answers1

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Clearly $x\neq 0$. So, your equation is equivalent to: $$ \sin x + \cos x = \frac{1}{x}$$ Graphing both LHS and RHS in the interval $x \in (0,2\pi)$:

We see that there are $3$ possible values of $x$ which satisfy the equation and hence there does exist values of $x$ for which your determinant is zero.

Note: You can use Newton's Method to approximate these values.

You'll find the values to be almost: $$x \approx 0.7, \quad x \approx 2, \quad x \approx 5.6$$ The above are extreme approximations and you can achieve better precision with more iterations

Nick
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