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Popular Trigonometry >

400+290cos(x)+290sin(x)=0

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Solution

400+290cos(x)+290sin(x)=0

Solution

x=−2.13356…+2πn,x=−2.57882…+2πn
+1
Degrees
x=−122.24412…∘+360∘n,x=−147.75587…∘+360∘n
Solution steps
400+290cos(x)+290sin(x)=0
Subtract 290sin(x) from both sides400+290cos(x)=−290sin(x)
Square both sides(400+290cos(x))2=(−290sin(x))2
Subtract (−290sin(x))2 from both sides(400+290cos(x))2−84100sin2(x)=0
Rewrite using trig identities
(400+290cos(x))2−84100sin2(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=(400+290cos(x))2−84100(1−cos2(x))
Simplify (400+290cos(x))2−84100(1−cos2(x)):168200cos2(x)+232000cos(x)+75900
(400+290cos(x))2−84100(1−cos2(x))
(400+290cos(x))2:160000+232000cos(x)+84100cos2(x)
Apply Perfect Square Formula: (a+b)2=a2+2ab+b2a=400,b=290cos(x)
=4002+2⋅400⋅290cos(x)+(290cos(x))2
Simplify 4002+2⋅400⋅290cos(x)+(290cos(x))2:160000+232000cos(x)+84100cos2(x)
4002+2⋅400⋅290cos(x)+(290cos(x))2
4002=160000
4002
4002=160000=160000
2⋅400⋅290cos(x)=232000cos(x)
2⋅400⋅290cos(x)
Multiply the numbers: 2⋅400⋅290=232000=232000cos(x)
(290cos(x))2=84100cos2(x)
(290cos(x))2
Apply exponent rule: (a⋅b)n=anbn=2902cos2(x)
2902=84100=84100cos2(x)
=160000+232000cos(x)+84100cos2(x)
=160000+232000cos(x)+84100cos2(x)
=160000+232000cos(x)+84100cos2(x)−84100(1−cos2(x))
Expand −84100(1−cos2(x)):−84100+84100cos2(x)
−84100(1−cos2(x))
Apply the distributive law: a(b−c)=ab−aca=−84100,b=1,c=cos2(x)=−84100⋅1−(−84100)cos2(x)
Apply minus-plus rules−(−a)=a=−84100⋅1+84100cos2(x)
Multiply the numbers: 84100⋅1=84100=−84100+84100cos2(x)
=160000+232000cos(x)+84100cos2(x)−84100+84100cos2(x)
Simplify 160000+232000cos(x)+84100cos2(x)−84100+84100cos2(x):168200cos2(x)+232000cos(x)+75900
160000+232000cos(x)+84100cos2(x)−84100+84100cos2(x)
Group like terms=232000cos(x)+84100cos2(x)+84100cos2(x)+160000−84100
Add similar elements: 84100cos2(x)+84100cos2(x)=168200cos2(x)=232000cos(x)+168200cos2(x)+160000−84100
Add/Subtract the numbers: 160000−84100=75900=168200cos2(x)+232000cos(x)+75900
=168200cos2(x)+232000cos(x)+75900
=168200cos2(x)+232000cos(x)+75900
75900+168200cos2(x)+232000cos(x)=0
Solve by substitution
75900+168200cos2(x)+232000cos(x)=0
Let: cos(x)=u75900+168200u2+232000u=0
75900+168200u2+232000u=0:u=58−40+82​​,u=58−40−82​​
75900+168200u2+232000u=0
Divide both sides by 16820016820075900​+168200168200u2​+168200232000u​=1682000​
Write in the standard form ax2+bx+c=0u2+2940u​+1682759​=0
Solve with the quadratic formula
u2+2940u​+1682759​=0
Quadratic Equation Formula:
For a=1,b=2940​,c=1682759​u1,2​=2⋅1−2940​±(2940​)2−4⋅1⋅1682759​​​
u1,2​=2⋅1−2940​±(2940​)2−4⋅1⋅1682759​​​
(2940​)2−4⋅1⋅1682759​​=2982​​
(2940​)2−4⋅1⋅1682759​​
(2940​)2=292402​
(2940​)2
Apply exponent rule: (ba​)c=bcac​=292402​
4⋅1⋅1682759​=8411518​
4⋅1⋅1682759​
Multiply fractions: a⋅cb​=ca⋅b​=1⋅1682759⋅4​
1682759⋅4​=8411518​
1682759⋅4​
Multiply the numbers: 759⋅4=3036=16823036​
Cancel the common factor: 2=8411518​
=1⋅8411518​
Multiply: 1⋅8411518​=8411518​=8411518​
=292402​−8411518​​
292402​=8411600​
292402​
402=1600=2921600​
292=841=8411600​
=8411600​−8411518​​
Combine the fractions 8411600​−8411518​:84182​
Apply rule ca​±cb​=ca±b​=8411600−1518​
Subtract the numbers: 1600−1518=82=84182​
=84182​​
Apply radical rule: assuming a≥0,b≥0=841​82​​
841​=29
841​
Factor the number: 841=292=292​
Apply radical rule: 292​=29=29
=2982​​
u1,2​=2⋅1−2940​±2982​​​
Separate the solutionsu1​=2⋅1−2940​+2982​​​,u2​=2⋅1−2940​−2982​​​
u=2⋅1−2940​+2982​​​:58−40+82​​
2⋅1−2940​+2982​​​
Combine the fractions −2940​+2982​​:29−40+82​​
Apply rule ca​±cb​=ca±b​=29−40+82​​
=2⋅129−40+82​​​
Multiply the numbers: 2⋅1=2=229−40+82​​​
Apply the fraction rule: acb​​=c⋅ab​=29⋅2−40+82​​
Multiply the numbers: 29⋅2=58=58−40+82​​
u=2⋅1−2940​−2982​​​:58−40−82​​
2⋅1−2940​−2982​​​
Combine the fractions −2940​−2982​​:29−40−82​​
Apply rule ca​±cb​=ca±b​=29−40−82​​
=2⋅129−40−82​​​
Multiply the numbers: 2⋅1=2=229−40−82​​​
Apply the fraction rule: acb​​=c⋅ab​=29⋅2−40−82​​
Multiply the numbers: 29⋅2=58=58−40−82​​
The solutions to the quadratic equation are:u=58−40+82​​,u=58−40−82​​
Substitute back u=cos(x)cos(x)=58−40+82​​,cos(x)=58−40−82​​
cos(x)=58−40+82​​,cos(x)=58−40−82​​
cos(x)=58−40+82​​:x=arccos(58−40+82​​)+2πn,x=−arccos(58−40+82​​)+2πn
cos(x)=58−40+82​​
Apply trig inverse properties
cos(x)=58−40+82​​
General solutions for cos(x)=58−40+82​​cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(58−40+82​​)+2πn,x=−arccos(58−40+82​​)+2πn
x=arccos(58−40+82​​)+2πn,x=−arccos(58−40+82​​)+2πn
cos(x)=58−40−82​​:x=arccos(58−40−82​​)+2πn,x=−arccos(58−40−82​​)+2πn
cos(x)=58−40−82​​
Apply trig inverse properties
cos(x)=58−40−82​​
General solutions for cos(x)=58−40−82​​cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(58−40−82​​)+2πn,x=−arccos(58−40−82​​)+2πn
x=arccos(58−40−82​​)+2πn,x=−arccos(58−40−82​​)+2πn
Combine all the solutionsx=arccos(58−40+82​​)+2πn,x=−arccos(58−40+82​​)+2πn,x=arccos(58−40−82​​)+2πn,x=−arccos(58−40−82​​)+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into 400+290cos(x)+290sin(x)=0
Remove the ones that don't agree with the equation.
Check the solution arccos(58−40+82​​)+2πn:False
arccos(58−40+82​​)+2πn
Plug in n=1arccos(58−40+82​​)+2π1
For 400+290cos(x)+290sin(x)=0plug inx=arccos(58−40+82​​)+2π1400+290cos(arccos(58−40+82​​)+2π1)+290sin(arccos(58−40+82​​)+2π1)=0
Refine490.55385…=0
⇒False
Check the solution −arccos(58−40+82​​)+2πn:True
−arccos(58−40+82​​)+2πn
Plug in n=1−arccos(58−40+82​​)+2π1
For 400+290cos(x)+290sin(x)=0plug inx=−arccos(58−40+82​​)+2π1400+290cos(−arccos(58−40+82​​)+2π1)+290sin(−arccos(58−40+82​​)+2π1)=0
Refine0=0
⇒True
Check the solution arccos(58−40−82​​)+2πn:False
arccos(58−40−82​​)+2πn
Plug in n=1arccos(58−40−82​​)+2π1
For 400+290cos(x)+290sin(x)=0plug inx=arccos(58−40−82​​)+2π1400+290cos(arccos(58−40−82​​)+2π1)+290sin(arccos(58−40−82​​)+2π1)=0
Refine309.44614…=0
⇒False
Check the solution −arccos(58−40−82​​)+2πn:True
−arccos(58−40−82​​)+2πn
Plug in n=1−arccos(58−40−82​​)+2π1
For 400+290cos(x)+290sin(x)=0plug inx=−arccos(58−40−82​​)+2π1400+290cos(−arccos(58−40−82​​)+2π1)+290sin(−arccos(58−40−82​​)+2π1)=0
Refine0=0
⇒True
x=−arccos(58−40+82​​)+2πn,x=−arccos(58−40−82​​)+2πn
Show solutions in decimal formx=−2.13356…+2πn,x=−2.57882…+2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for 400+290cos(x)+290sin(x)=0 ?

    The general solution for 400+290cos(x)+290sin(x)=0 is x=-2.13356…+2pin,x=-2.57882…+2pin
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