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L'Hôpital's rule: limit at 0 example (video) | Khan Academy

Sal uses L'Hôpital's rule to **find** the limit at 0 of (2sin(x)-**sin**(**2x**))/(x-**sin**(**x**)). ... Let's
say we need to evaluate the limit as x approaches 0 of 2 sine of x minus sine of ...

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If **sin** (**x**-**20**)° = **cos** (**3x**-**10**)°, what is the value of x? - Quora

I am interpreting the question as: Let x be an angle expressed in degrees, and let
**sin**(**x**-**20**) = **3x** -**10**. **Find** x. I ...

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**Solving** Simple (to Medium-Hard) **Trig Equations** | Purplemath

**Solving trig equations** use both the reference angles and **trigonometric identities**
that you've memorized, together ... between radians and degrees, and what the
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If **sin**(**2x** + **10**)=**cos**(**3x**-**20**) , **find** the value of x , in the book says that ...

May 12, 2014 **...** If **sin**(**2x** + **10**)=**cos**(**3x**-**20**) , **find** the value of x , in the book says that the x is 1/4
but i dont know how to come to this? Thank You?

For more information, see If **sin**(**2x** + **10**)=**cos**(**3x**-**20**) , **find** the value of x , in the book says that ...

We want to find the value of x for which sin(x+10) = cos(3x+20) where 0 <= x <= 360.

We have the following relationships between sin and cos,

sin(y+90) = cos(y) and sin(90-y) = cos(y).

So, the above equations can be rewritten as either.

sin(x+10) = sin(3x+20+90) => sin(x+10) = sin(3x+110) - (1)

or

siin(x+10) = sin(90 - 3x - 20) => sin(x+10) = sin(70 - 3x) - (2)

We know that if sin(a) = sin(b), then a = b +/- n*360 (n = 0, 1, 2, 3, ...), or a = 180 - b +/- n*360 (n = 0, 1, 2, 3, ...)

From equation 1, we have

x + 10 = 3x + 110 +/- n*360 => x = -50 +/- n*180 => x = 130 +/- n*180

or

x + 10 = 180 - (3x + 110) +/- n*360 => x = 60/4 +/- n*90 = 15 +/- n*90

From equation 2, we have

x + 10 = 70 - 3x +/- n*360 => x = 60/4 +/ n*90 = 15 +/- n*90

or

x + 10 = 180 - (70 - 3x) +/- n*360 => x = -50 +/- n*180 = 130 +/- n*180

So, the possible solutions for x are:

x = 15 +/- n*90 (n = 0, 1, 2, 3, ...)

x = 130 +/- n*180 (n = 0, 1, 2, 3, ...)

So, all possible unique solutions for 0 <= x <= 360

x = 15, 105, 195, 285

x = 130, 310

You can verify the answers by plugging in the values into the original equation (use wolframalpha.com) and checking that both the left and right hand sides evaluate to the same value.