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Popular Trigonometry >

5*cos(x)+4-2*sin(x)=0

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Solution

5⋅cos(x)+4−2⋅sin(x)=0

Solution

x=2.02750…+2πn,x=−2.78851…+2πn
+1
Degrees
x=116.16747…∘+360∘n,x=−159.77029…∘+360∘n
Solution steps
5cos(x)+4−2sin(x)=0
Add 2sin(x) to both sides5cos(x)+4=2sin(x)
Square both sides(5cos(x)+4)2=(2sin(x))2
Subtract (2sin(x))2 from both sides(5cos(x)+4)2−4sin2(x)=0
Rewrite using trig identities
(4+5cos(x))2−4sin2(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=(4+5cos(x))2−4(1−cos2(x))
Simplify (4+5cos(x))2−4(1−cos2(x)):29cos2(x)+40cos(x)+12
(4+5cos(x))2−4(1−cos2(x))
(4+5cos(x))2:16+40cos(x)+25cos2(x)
Apply Perfect Square Formula: (a+b)2=a2+2ab+b2a=4,b=5cos(x)
=42+2⋅4⋅5cos(x)+(5cos(x))2
Simplify 42+2⋅4⋅5cos(x)+(5cos(x))2:16+40cos(x)+25cos2(x)
42+2⋅4⋅5cos(x)+(5cos(x))2
42=16
42
42=16=16
2⋅4⋅5cos(x)=40cos(x)
2⋅4⋅5cos(x)
Multiply the numbers: 2⋅4⋅5=40=40cos(x)
(5cos(x))2=25cos2(x)
(5cos(x))2
Apply exponent rule: (a⋅b)n=anbn=52cos2(x)
52=25=25cos2(x)
=16+40cos(x)+25cos2(x)
=16+40cos(x)+25cos2(x)
=16+40cos(x)+25cos2(x)−4(1−cos2(x))
Expand −4(1−cos2(x)):−4+4cos2(x)
−4(1−cos2(x))
Apply the distributive law: a(b−c)=ab−aca=−4,b=1,c=cos2(x)=−4⋅1−(−4)cos2(x)
Apply minus-plus rules−(−a)=a=−4⋅1+4cos2(x)
Multiply the numbers: 4⋅1=4=−4+4cos2(x)
=16+40cos(x)+25cos2(x)−4+4cos2(x)
Simplify 16+40cos(x)+25cos2(x)−4+4cos2(x):29cos2(x)+40cos(x)+12
16+40cos(x)+25cos2(x)−4+4cos2(x)
Group like terms=40cos(x)+25cos2(x)+4cos2(x)+16−4
Add similar elements: 25cos2(x)+4cos2(x)=29cos2(x)=40cos(x)+29cos2(x)+16−4
Add/Subtract the numbers: 16−4=12=29cos2(x)+40cos(x)+12
=29cos2(x)+40cos(x)+12
=29cos2(x)+40cos(x)+12
12+29cos2(x)+40cos(x)=0
Solve by substitution
12+29cos2(x)+40cos(x)=0
Let: cos(x)=u12+29u2+40u=0
12+29u2+40u=0:u=292(13​−10)​,u=−292(10+13​)​
12+29u2+40u=0
Write in the standard form ax2+bx+c=029u2+40u+12=0
Solve with the quadratic formula
29u2+40u+12=0
Quadratic Equation Formula:
For a=29,b=40,c=12u1,2​=2⋅29−40±402−4⋅29⋅12​​
u1,2​=2⋅29−40±402−4⋅29⋅12​​
402−4⋅29⋅12​=413​
402−4⋅29⋅12​
Multiply the numbers: 4⋅29⋅12=1392=402−1392​
402=1600=1600−1392​
Subtract the numbers: 1600−1392=208=208​
Prime factorization of 208:24⋅13
208
208divides by 2208=104⋅2=2⋅104
104divides by 2104=52⋅2=2⋅2⋅52
52divides by 252=26⋅2=2⋅2⋅2⋅26
26divides by 226=13⋅2=2⋅2⋅2⋅2⋅13
2,13 are all prime numbers, therefore no further factorization is possible=2⋅2⋅2⋅2⋅13
=24⋅13
=24⋅13​
Apply radical rule: nab​=na​nb​=13​24​
Apply radical rule: nam​=anm​24​=224​=22=2213​
Refine=413​
u1,2​=2⋅29−40±413​​
Separate the solutionsu1​=2⋅29−40+413​​,u2​=2⋅29−40−413​​
u=2⋅29−40+413​​:292(13​−10)​
2⋅29−40+413​​
Multiply the numbers: 2⋅29=58=58−40+413​​
Factor −40+413​:4(−10+13​)
−40+413​
Rewrite as=−4⋅10+413​
Factor out common term 4=4(−10+13​)
=584(−10+13​)​
Cancel the common factor: 2=292(13​−10)​
u=2⋅29−40−413​​:−292(10+13​)​
2⋅29−40−413​​
Multiply the numbers: 2⋅29=58=58−40−413​​
Factor −40−413​:−4(10+13​)
−40−413​
Rewrite as=−4⋅10−413​
Factor out common term 4=−4(10+13​)
=−584(10+13​)​
Cancel the common factor: 2=−292(10+13​)​
The solutions to the quadratic equation are:u=292(13​−10)​,u=−292(10+13​)​
Substitute back u=cos(x)cos(x)=292(13​−10)​,cos(x)=−292(10+13​)​
cos(x)=292(13​−10)​,cos(x)=−292(10+13​)​
cos(x)=292(13​−10)​:x=arccos(292(13​−10)​)+2πn,x=−arccos(292(13​−10)​)+2πn
cos(x)=292(13​−10)​
Apply trig inverse properties
cos(x)=292(13​−10)​
General solutions for cos(x)=292(13​−10)​cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(292(13​−10)​)+2πn,x=−arccos(292(13​−10)​)+2πn
x=arccos(292(13​−10)​)+2πn,x=−arccos(292(13​−10)​)+2πn
cos(x)=−292(10+13​)​:x=arccos(−292(10+13​)​)+2πn,x=−arccos(−292(10+13​)​)+2πn
cos(x)=−292(10+13​)​
Apply trig inverse properties
cos(x)=−292(10+13​)​
General solutions for cos(x)=−292(10+13​)​cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−292(10+13​)​)+2πn,x=−arccos(−292(10+13​)​)+2πn
x=arccos(−292(10+13​)​)+2πn,x=−arccos(−292(10+13​)​)+2πn
Combine all the solutionsx=arccos(292(13​−10)​)+2πn,x=−arccos(292(13​−10)​)+2πn,x=arccos(−292(10+13​)​)+2πn,x=−arccos(−292(10+13​)​)+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into 5cos(x)+4−2sin(x)=0
Remove the ones that don't agree with the equation.
Check the solution arccos(292(13​−10)​)+2πn:True
arccos(292(13​−10)​)+2πn
Plug in n=1arccos(292(13​−10)​)+2π1
For 5cos(x)+4−2sin(x)=0plug inx=arccos(292(13​−10)​)+2π15cos(arccos(292(13​−10)​)+2π1)+4−2sin(arccos(292(13​−10)​)+2π1)=0
Refine0=0
⇒True
Check the solution −arccos(292(13​−10)​)+2πn:False
−arccos(292(13​−10)​)+2πn
Plug in n=1−arccos(292(13​−10)​)+2π1
For 5cos(x)+4−2sin(x)=0plug inx=−arccos(292(13​−10)​)+2π15cos(−arccos(292(13​−10)​)+2π1)+4−2sin(−arccos(292(13​−10)​)+2π1)=0
Refine3.59003…=0
⇒False
Check the solution arccos(−292(10+13​)​)+2πn:False
arccos(−292(10+13​)​)+2πn
Plug in n=1arccos(−292(10+13​)​)+2π1
For 5cos(x)+4−2sin(x)=0plug inx=arccos(−292(10+13​)​)+2π15cos(arccos(−292(10+13​)​)+2π1)+4−2sin(arccos(−292(10+13​)​)+2π1)=0
Refine−1.38313…=0
⇒False
Check the solution −arccos(−292(10+13​)​)+2πn:True
−arccos(−292(10+13​)​)+2πn
Plug in n=1−arccos(−292(10+13​)​)+2π1
For 5cos(x)+4−2sin(x)=0plug inx=−arccos(−292(10+13​)​)+2π15cos(−arccos(−292(10+13​)​)+2π1)+4−2sin(−arccos(−292(10+13​)​)+2π1)=0
Refine0=0
⇒True
x=arccos(292(13​−10)​)+2πn,x=−arccos(−292(10+13​)​)+2πn
Show solutions in decimal formx=2.02750…+2πn,x=−2.78851…+2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for 5*cos(x)+4-2*sin(x)=0 ?

    The general solution for 5*cos(x)+4-2*sin(x)=0 is x=2.02750…+2pin,x=-2.78851…+2pin
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