Trigonometry Examples

Solve for x sin((7pi)/6)+sin(2((7pi)/6))=(sin(x)+sin(2x)((7pi)/6))
sin(7π6)+sin(2(7π6))=(sin(x)+sin(2x)(7π6))sin(7π6)+sin(2(7π6))=(sin(x)+sin(2x)(7π6))
Step 1
Simplify sin(7π6)+sin(2(7π6))sin(7π6)+sin(2(7π6)).
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Step 1.1
Remove parentheses.
sin(7π6)+sin(2(7π6))=sin(x)+sin(2x)(7π6)sin(7π6)+sin(2(7π6))=sin(x)+sin(2x)(7π6)
Step 1.2
Simplify each term.
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Step 1.2.1
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the third quadrant.
-sin(π6)+sin(2(7π6))=sin(x)+sin(2x)(7π6)sin(π6)+sin(2(7π6))=sin(x)+sin(2x)(7π6)
Step 1.2.2
The exact value of sin(π6)sin(π6) is 1212.
-12+sin(2(7π6))=sin(x)+sin(2x)(7π6)12+sin(2(7π6))=sin(x)+sin(2x)(7π6)
Step 1.2.3
Cancel the common factor of 22.
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Step 1.2.3.1
Factor 22 out of 66.
-12+sin(27π2(3))=sin(x)+sin(2x)(7π6)12+sin(27π2(3))=sin(x)+sin(2x)(7π6)
Step 1.2.3.2
Cancel the common factor.
-12+sin(27π23)=sin(x)+sin(2x)(7π6)
Step 1.2.3.3
Rewrite the expression.
-12+sin(7π3)=sin(x)+sin(2x)(7π6)
-12+sin(7π3)=sin(x)+sin(2x)(7π6)
Step 1.2.4
Subtract full rotations of 2π until the angle is greater than or equal to 0 and less than 2π.
-12+sin(π3)=sin(x)+sin(2x)(7π6)
Step 1.2.5
The exact value of sin(π3) is 32.
-12+32=sin(x)+sin(2x)(7π6)
-12+32=sin(x)+sin(2x)(7π6)
-12+32=sin(x)+sin(2x)(7π6)
Step 2
Simplify sin(x)+sin(2x)(7π6).
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Step 2.1
Simplify each term.
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Step 2.1.1
Combine sin(2x) and 7π6.
-12+32=sin(x)+sin(2x)(7π)6
Step 2.1.2
Move 7 to the left of sin(2x).
-12+32=sin(x)+7sin(2x)π6
-12+32=sin(x)+7sin(2x)π6
Step 2.2
Reorder factors in sin(x)+7sin(2x)π6.
-12+32=sin(x)+7πsin(2x)6
-12+32=sin(x)+7πsin(2x)6
Step 3
Graph each side of the equation. The solution is the x-value of the point of intersection.
x0.04399022+2πn,1.6572027+2πn,3.1995582+2πn,4.52402682+2πn, for any integer n
Step 4
 [x2  12  π  xdx ]