How do you solve sin(x+10) = cos(3x+20).

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