1. ## Solving trig equation

Find solutions:

sin ( $\displaystyle \frac{1}{2}$ - 7x) $\displaystyle \times$ $\displaystyle \pi$ = cos(2-5x) $\displaystyle \times$ $\displaystyle \pi$ + sin8$\displaystyle \pi$

Interval - [$\displaystyle \frac{1}{2}$;1]
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What I did: 1) sin ( $\displaystyle \frac{1}{2}$ - 7x) = sin $\displaystyle \frac{1}{2}$cos7x + sin(7x)cos($\displaystyle \frac{1}{2}$)

2) cos(2-5x) = cos2cos5x + sin2sin5x

..and 3) sin8$\displaystyle \pi$ = sin2$\displaystyle \pi$

But what should I do now, should I multiply by $\displaystyle \pi$? Thank you.

2. Do you mean $\displaystyle \sin \left(\dfrac{1}{2}-7x\right) \times \pi$ or $\displaystyle \sin \left[\left \pi (\dfrac{1}{2}-7x\right)\right]$

Same for the other terms.

3. Originally Posted by e^(i*pi)
Do you mean $\displaystyle \sin \left(\dfrac{1}{2}-7x\right) \times \pi$ or $\displaystyle \sin \left[\left \pi (\dfrac{1}{2}-7x\right)\right]$

Same for the other terms.
Sorry, it is $\displaystyle \sin \left(\dfrac{1}{2}-7x\right) \times \pi$

4. Hi!

$\displaystyle \pi\cdot\sin{\left(\frac{1}{2} - 7x\right)} = \pi\cdot\cos{\left(2-5x\right)} + \sin{8\pi}$

Now, considering that $\displaystyle \sin{8\pi}=0$ and that the 'pi'-es on both sides cancel each other out, the equation turns to:

$\displaystyle \sin{\left(\frac{1}{2} - 7x\right)} = \cos{\left(\2-5x\right)}$

By squaring both sides, we'll be able to use an identity that'll get rid of the unsightly fraction in the sine:

$\displaystyle \sin^2{\left(\frac{1}{2} - 7x\right)} = \cos^2{\left(2-5x\right)}$
$\displaystyle \frac{1 - \cos{(1-14x)}}{2} = \frac{1 + \cos{(4-10x)}}{2}$
$\displaystyle -\cos{(1-14x)} = \cos{(4-10x)}$
$\displaystyle -\cos{(1-14x)}-\cos{(4-10x)} = 0$

And after multiplying it by (-1), we convert it from a sum to a product:

$\displaystyle 2\cos{\left(\frac{1-14x+4-10x}{2}\right)}\cos{\left(\frac{1-14x-4x+10x}{2}\right)} = 0$

Which is equivalent to:

$\displaystyle \cos{\left(\frac{5}{2} - 12x\right)}\cos{\left(-\frac{3}{2} - 2x\right)} = 0$,

and by equating each factor with zero, you get the solutions:

$\displaystyle x_1 = \frac{5-\pi}{24} + k\cdot\frac{\pi}{6}, \, k\in\mathbf{Z}$
$\displaystyle x_2 = \frac{-3-\pi}{4} + k\cdot\pi, \, k\in\mathbf{Z}$

We find the solution within the interval $\displaystyle \left[\frac{1}{2}, 1\right]$ by including arbitrary values of the integer 'k'. By including k=1, we find the only solution, derived from the first general solution, and that is:

$\displaystyle x = \frac{5+3\pi}{24},\, k=1$

Hope I helped!

5. Originally Posted by mrfour44
$\displaystyle \sin{\left(\frac{1}{2} - 7x\right)} = \cos{\left(\2-5x\right)}$
At this stage, instead of squaring you can write:

$\displaystyle \cos\left(\frac{\pi}{2}-\frac{1}{2}+7x\right) = \cos \left( 2-5x \right)$

Now the general solution for the cosine is:

$\displaystyle \cos{\theta} = \cos{\alpha} \Leftrightarrow \theta = 2n\pi \pm \alpha.$

From this it of course follows that

$\displaystyle x = \left(4n-1\right)\frac{\pi}{4}-\frac{3}{4}, ~~ \left(4n-1\right)\frac{\pi}{24}+\frac{5}{24}.$

6. That's so much easier. Forget mine