Question 1203265: Cos(x+35)=sin(2x+10). Find the value of x
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13367)   (Show Source):
You can put this solution on YOUR website!
There is one simple solution and two others that require more work to find.
(1) If the two angles are both in quadrant I, then the sine of one is equal to the cosine of the other if the sum of the two angle measures is 90 degrees.
(x+35)+(2x+10) = 90
3x+45 = 90
3x = 45
x = 15
ANSWER: x=15; the two angle measures are 50 and 40 (sum 90).
The other solutions are when the angles are in other quadrants. In those cases, the sine of one is equal to the cosine of the other if the sum of the measures of the REFERENCE angles of the two angles is 90 degrees.
(2) There could be a solution with x+35 in quadrant II (cosine negative) and 2x+10 in quadrant III (sine negative).
For an angle of x+35 degrees in quadrant II, the reference angle is 180-(x+35) = 145-x; for an angle of 2x+10 degrees in quadrant III, the reference angle is (2x+10)-180 = 2x-170.
(145-x)+(2x-170) = 90
x-25 = 90
x = 115
ANSWER: x=115; the two angle measures are 150 and 240; the reference angles are 30 and 60 (sum 90).
(3) There could be a solution with x+35 in quadrant II (cosine negative) and 2x+10 in quadrant IV (sine negative).
As in case (2), the reference angle for an angle of x+35 in quadrant II is 145-x; for an angle of 2x+10 degrees in quadrant IV, the reference angle is 360-(2x+10) = 350-2x.
(145-x)+(350-2x) = 90
495-3x = 90
3x = 405
x = 135
ANSWER: x=135; the two angle measures are 170 and 280; the reference angles are 10 and 80 (sum 90).
A graph shows the three solutions at x=15, x=115, and x=135:
Note that because of the periodicity of the graphs, there are an infinite number of other solutions; the three shown are the only ones for which x+35 and 2x+10 are both between 0 and 360 degrees.
Answer by ikleyn(53937)  (Show Source):
You can put this solution on YOUR website! .
Cos(x+35)=sin(2x+10). Find the value of x
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To the problem's composer.
As usually at this forum, the problem is posed, worded, printed and presented incorrectly,
because in this formulation, the number of possible answers/solutions in INFINITELY LARGE.
To be as accurate as it is required, the problem must define the range (the diapason) of x's,
for which we are looking the solutions: for example, 0 <= x < 360 degrees, as a typical restriction.
Otherwise, mathematical and grammatical incompetence of the Math composer becomes visible
to everyone, even with unarmed eyes.
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Do not rush to post your anathema to me in response.
You better fix your writing and post your "THANKS" to me for my teaching.
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