Trigonometry Examples

Solve for C in Degrees -5cos(C)-1=2cos(C)+3
-5cos(C)-1=2cos(C)+35cos(C)1=2cos(C)+3
Step 1
Move all terms containing cos(C)cos(C) to the left side of the equation.
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Step 1.1
Subtract 2cos(C)2cos(C) from both sides of the equation.
-5cos(C)-1-2cos(C)=35cos(C)12cos(C)=3
Step 1.2
Subtract 2cos(C)2cos(C) from -5cos(C)5cos(C).
-7cos(C)-1=37cos(C)1=3
-7cos(C)-1=37cos(C)1=3
Step 2
Move all terms not containing cos(C)cos(C) to the right side of the equation.
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Step 2.1
Add 11 to both sides of the equation.
-7cos(C)=3+17cos(C)=3+1
Step 2.2
Add 33 and 11.
-7cos(C)=47cos(C)=4
-7cos(C)=47cos(C)=4
Step 3
Divide each term in -7cos(C)=47cos(C)=4 by -77 and simplify.
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Step 3.1
Divide each term in -7cos(C)=47cos(C)=4 by -77.
-7cos(C)-7=4-77cos(C)7=47
Step 3.2
Simplify the left side.
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Step 3.2.1
Cancel the common factor of -77.
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Step 3.2.1.1
Cancel the common factor.
-7cos(C)-7=4-7
Step 3.2.1.2
Divide cos(C) by 1.
cos(C)=4-7
cos(C)=4-7
cos(C)=4-7
Step 3.3
Simplify the right side.
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Step 3.3.1
Move the negative in front of the fraction.
cos(C)=-47
cos(C)=-47
cos(C)=-47
Step 4
Take the inverse cosine of both sides of the equation to extract C from inside the cosine.
C=arccos(-47)
Step 5
Simplify the right side.
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Step 5.1
Evaluate arccos(-47).
C=124.84990457
C=124.84990457
Step 6
The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from 360 to find the solution in the third quadrant.
C=360-124.84990457
Step 7
Subtract 124.84990457 from 360.
C=235.15009542
Step 8
Find the period of cos(C).
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Step 8.1
The period of the function can be calculated using 360|b|.
360|b|
Step 8.2
Replace b with 1 in the formula for period.
360|1|
Step 8.3
The absolute value is the distance between a number and zero. The distance between 0 and 1 is 1.
3601
Step 8.4
Divide 360 by 1.
360
360
Step 9
The period of the cos(C) function is 360 so values will repeat every 360 degrees in both directions.
C=124.84990457+360n,235.15009542+360n, for any integer n
 [x2  12  π  xdx ]