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Popular Trigonometry >

2sin(x)+5cos(x)=4

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Solution

2sin(x)+5cos(x)=4

Solution

x=1.11408…+2πn,x=2π−0.35307…+2πn
+1
Degrees
x=63.83252…∘+360∘n,x=339.77029…∘+360∘n
Solution steps
2sin(x)+5cos(x)=4
Subtract 5cos(x) from both sides2sin(x)=4−5cos(x)
Square both sides(2sin(x))2=(4−5cos(x))2
Subtract (4−5cos(x))2 from both sides4sin2(x)−16+40cos(x)−25cos2(x)=0
Rewrite using trig identities
−16−25cos2(x)+40cos(x)+4sin2(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−16−25cos2(x)+40cos(x)+4(1−cos2(x))
Simplify −16−25cos2(x)+40cos(x)+4(1−cos2(x)):40cos(x)−29cos2(x)−12
−16−25cos2(x)+40cos(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⋅1−4cos2(x)
Multiply the numbers: 4⋅1=4=4−4cos2(x)
=−16−25cos2(x)+40cos(x)+4−4cos2(x)
Simplify −16−25cos2(x)+40cos(x)+4−4cos2(x):40cos(x)−29cos2(x)−12
−16−25cos2(x)+40cos(x)+4−4cos2(x)
Group like terms=−25cos2(x)+40cos(x)−4cos2(x)−16+4
Add similar elements: −25cos2(x)−4cos2(x)=−29cos2(x)=−29cos2(x)+40cos(x)−16+4
Add/Subtract the numbers: −16+4=−12=40cos(x)−29cos2(x)−12
=40cos(x)−29cos2(x)−12
=40cos(x)−29cos2(x)−12
−12−29cos2(x)+40cos(x)=0
Solve by substitution
−12−29cos2(x)+40cos(x)=0
Let: cos(x)=u−12−29u2+40u=0
−12−29u2+40u=0:u=292(10−13​)​,u=292(10+13​)​
−12−29u2+40u=0
Write in the standard form ax2+bx+c=0−29u2+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)​
Apply rule −(−a)=a=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: =13​24​
Apply radical rule: 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(10−13​)​
2(−29)−40+413​​
Remove parentheses: (−a)=−a=−2⋅29−40+413​​
Multiply the numbers: 2⋅29=58=−58−40+413​​
Apply the fraction rule: −b−a​=ba​−40+413​=−(40−413​)=5840−413​​
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​)​
u=2(−29)−40−413​​:292(10+13​)​
2(−29)−40−413​​
Remove parentheses: (−a)=−a=−2⋅29−40−413​​
Multiply the numbers: 2⋅29=58=−58−40−413​​
Apply the fraction rule: −b−a​=ba​−40−413​=−(40+413​)=5840+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(10−13​)​,u=292(10+13​)​
Substitute back u=cos(x)cos(x)=292(10−13​)​,cos(x)=292(10+13​)​
cos(x)=292(10−13​)​,cos(x)=292(10+13​)​
cos(x)=292(10−13​)​:x=arccos(292(10−13​)​)+2πn,x=2π−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=2π−arccos(a)+2πnx=arccos(292(10−13​)​)+2πn,x=2π−arccos(292(10−13​)​)+2πn
x=arccos(292(10−13​)​)+2πn,x=2π−arccos(292(10−13​)​)+2πn
cos(x)=292(10+13​)​:x=arccos(292(10+13​)​)+2πn,x=2π−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=2π−arccos(a)+2πnx=arccos(292(10+13​)​)+2πn,x=2π−arccos(292(10+13​)​)+2πn
x=arccos(292(10+13​)​)+2πn,x=2π−arccos(292(10+13​)​)+2πn
Combine all the solutionsx=arccos(292(10−13​)​)+2πn,x=2π−arccos(292(10−13​)​)+2πn,x=arccos(292(10+13​)​)+2πn,x=2π−arccos(292(10+13​)​)+2πn
Verify solutions by plugging them into the original equation
Check the solutions by plugging them into 2sin(x)+5cos(x)=4
Remove the ones that don't agree with the equation.
Check the solution arccos(292(10−13​)​)+2πn:True
arccos(292(10−13​)​)+2πn
Plug in n=1arccos(292(10−13​)​)+2π1
For 2sin(x)+5cos(x)=4plug inx=arccos(292(10−13​)​)+2π12sin(arccos(292(10−13​)​)+2π1)+5cos(arccos(292(10−13​)​)+2π1)=4
Refine4=4
⇒True
Check the solution 2π−arccos(292(10−13​)​)+2πn:False
2π−arccos(292(10−13​)​)+2πn
Plug in n=12π−arccos(292(10−13​)​)+2π1
For 2sin(x)+5cos(x)=4plug inx=2π−arccos(292(10−13​)​)+2π12sin(2π−arccos(292(10−13​)​)+2π1)+5cos(2π−arccos(292(10−13​)​)+2π1)=4
Refine0.40996…=4
⇒False
Check the solution arccos(292(10+13​)​)+2πn:False
arccos(292(10+13​)​)+2πn
Plug in n=1arccos(292(10+13​)​)+2π1
For 2sin(x)+5cos(x)=4plug inx=arccos(292(10+13​)​)+2π12sin(arccos(292(10+13​)​)+2π1)+5cos(arccos(292(10+13​)​)+2π1)=4
Refine5.38313…=4
⇒False
Check the solution 2π−arccos(292(10+13​)​)+2πn:True
2π−arccos(292(10+13​)​)+2πn
Plug in n=12π−arccos(292(10+13​)​)+2π1
For 2sin(x)+5cos(x)=4plug inx=2π−arccos(292(10+13​)​)+2π12sin(2π−arccos(292(10+13​)​)+2π1)+5cos(2π−arccos(292(10+13​)​)+2π1)=4
Refine4=4
⇒True
x=arccos(292(10−13​)​)+2πn,x=2π−arccos(292(10+13​)​)+2πn
Show solutions in decimal formx=1.11408…+2πn,x=2π−0.35307…+2πn

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Frequently Asked Questions (FAQ)

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

    The general solution for 2sin(x)+5cos(x)=4 is x=1.11408…+2pin,x=2pi-0.35307…+2pin
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