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Popular Trigonometry >

25cos^2(x)=9

  • Pre Algebra
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Solution

25cos2(x)=9

Solution

x=0.92729…+2πn,x=2π−0.92729…+2πn,x=2.21429…+2πn,x=−2.21429…+2πn
+1
Degrees
x=53.13010…∘+360∘n,x=306.86989…∘+360∘n,x=126.86989…∘+360∘n,x=−126.86989…∘+360∘n
Solution steps
25cos2(x)=9
Solve by substitution
25cos2(x)=9
Let: cos(x)=u25u2=9
25u2=9:u=53​,u=−53​
25u2=9
Divide both sides by 25
25u2=9
Divide both sides by 252525u2​=259​
Simplifyu2=259​
u2=259​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=259​​,u=−259​​
259​​=53​
259​​
Apply radical rule: nba​​=nb​na​​, assuming a≥0,b≥0=25​9​​
25​=5
25​
Factor the number: 25=52=52​
Apply radical rule: nan​=a52​=5=5
=59​​
9​=3
9​
Factor the number: 9=32=32​
Apply radical rule: nan​=a32​=3=3
=53​
−259​​=−53​
−259​​
Simplify 259​​:53​
259​​
Apply radical rule: nba​​=nb​na​​, assuming a≥0,b≥0=25​9​​
25​=5
25​
Factor the number: 25=52=52​
Apply radical rule: nan​=a52​=5=5
=59​​
9​=3
9​
Factor the number: 9=32=32​
Apply radical rule: nan​=a32​=3=3
=53​
=−53​
u=53​,u=−53​
Substitute back u=cos(x)cos(x)=53​,cos(x)=−53​
cos(x)=53​,cos(x)=−53​
cos(x)=53​:x=arccos(53​)+2πn,x=2π−arccos(53​)+2πn
cos(x)=53​
Apply trig inverse properties
cos(x)=53​
General solutions for cos(x)=53​cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos(53​)+2πn,x=2π−arccos(53​)+2πn
x=arccos(53​)+2πn,x=2π−arccos(53​)+2πn
cos(x)=−53​:x=arccos(−53​)+2πn,x=−arccos(−53​)+2πn
cos(x)=−53​
Apply trig inverse properties
cos(x)=−53​
General solutions for cos(x)=−53​cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos(−53​)+2πn,x=−arccos(−53​)+2πn
x=arccos(−53​)+2πn,x=−arccos(−53​)+2πn
Combine all the solutionsx=arccos(53​)+2πn,x=2π−arccos(53​)+2πn,x=arccos(−53​)+2πn,x=−arccos(−53​)+2πn
Show solutions in decimal formx=0.92729…+2πn,x=2π−0.92729…+2πn,x=2.21429…+2πn,x=−2.21429…+2πn

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