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Popular Trigonometry >

2sin^2(3x)-3cos(3x)+1=0

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Solution

2sin2(3x)−3cos(3x)+1=0

Solution

x=30.81462…​+32πn​,x=32π​−30.81462…​+32πn​
+1
Degrees
x=15.55820…∘+120∘n,x=104.44179…∘+120∘n
Solution steps
2sin2(3x)−3cos(3x)+1=0
Rewrite using trig identities
1+2sin2(3x)−3cos(3x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=1+2(1−cos2(3x))−3cos(3x)
Simplify 1+2(1−cos2(3x))−3cos(3x):−2cos2(3x)−3cos(3x)+3
1+2(1−cos2(3x))−3cos(3x)
Expand 2(1−cos2(3x)):2−2cos2(3x)
2(1−cos2(3x))
Apply the distributive law: a(b−c)=ab−aca=2,b=1,c=cos2(3x)=2⋅1−2cos2(3x)
Multiply the numbers: 2⋅1=2=2−2cos2(3x)
=1+2−2cos2(3x)−3cos(3x)
Add the numbers: 1+2=3=−2cos2(3x)−3cos(3x)+3
=−2cos2(3x)−3cos(3x)+3
3−2cos2(3x)−3cos(3x)=0
Solve by substitution
3−2cos2(3x)−3cos(3x)=0
Let: cos(3x)=u3−2u2−3u=0
3−2u2−3u=0:u=−43+33​​,u=433​−3​
3−2u2−3u=0
Write in the standard form ax2+bx+c=0−2u2−3u+3=0
Solve with the quadratic formula
−2u2−3u+3=0
Quadratic Equation Formula:
For a=−2,b=−3,c=3u1,2​=2(−2)−(−3)±(−3)2−4(−2)⋅3​​
u1,2​=2(−2)−(−3)±(−3)2−4(−2)⋅3​​
(−3)2−4(−2)⋅3​=33​
(−3)2−4(−2)⋅3​
Apply rule −(−a)=a=(−3)2+4⋅2⋅3​
Apply exponent rule: (−a)n=an,if n is even(−3)2=32=32+4⋅2⋅3​
Multiply the numbers: 4⋅2⋅3=24=32+24​
32=9=9+24​
Add the numbers: 9+24=33=33​
u1,2​=2(−2)−(−3)±33​​
Separate the solutionsu1​=2(−2)−(−3)+33​​,u2​=2(−2)−(−3)−33​​
u=2(−2)−(−3)+33​​:−43+33​​
2(−2)−(−3)+33​​
Remove parentheses: (−a)=−a,−(−a)=a=−2⋅23+33​​
Multiply the numbers: 2⋅2=4=−43+33​​
Apply the fraction rule: −ba​=−ba​=−43+33​​
u=2(−2)−(−3)−33​​:433​−3​
2(−2)−(−3)−33​​
Remove parentheses: (−a)=−a,−(−a)=a=−2⋅23−33​​
Multiply the numbers: 2⋅2=4=−43−33​​
Apply the fraction rule: −b−a​=ba​3−33​=−(33​−3)=433​−3​
The solutions to the quadratic equation are:u=−43+33​​,u=433​−3​
Substitute back u=cos(3x)cos(3x)=−43+33​​,cos(3x)=433​−3​
cos(3x)=−43+33​​,cos(3x)=433​−3​
cos(3x)=−43+33​​:No Solution
cos(3x)=−43+33​​
−1≤cos(x)≤1NoSolution
cos(3x)=433​−3​:x=3arccos(433​−3​)​+32πn​,x=32π​−3arccos(433​−3​)​+32πn​
cos(3x)=433​−3​
Apply trig inverse properties
cos(3x)=433​−3​
General solutions for cos(3x)=433​−3​cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πn3x=arccos(433​−3​)+2πn,3x=2π−arccos(433​−3​)+2πn
3x=arccos(433​−3​)+2πn,3x=2π−arccos(433​−3​)+2πn
Solve 3x=arccos(433​−3​)+2πn:x=3arccos(433​−3​)​+32πn​
3x=arccos(433​−3​)+2πn
Divide both sides by 3
3x=arccos(433​−3​)+2πn
Divide both sides by 333x​=3arccos(433​−3​)​+32πn​
Simplifyx=3arccos(433​−3​)​+32πn​
x=3arccos(433​−3​)​+32πn​
Solve 3x=2π−arccos(433​−3​)+2πn:x=32π​−3arccos(433​−3​)​+32πn​
3x=2π−arccos(433​−3​)+2πn
Divide both sides by 3
3x=2π−arccos(433​−3​)+2πn
Divide both sides by 333x​=32π​−3arccos(433​−3​)​+32πn​
Simplifyx=32π​−3arccos(433​−3​)​+32πn​
x=32π​−3arccos(433​−3​)​+32πn​
x=3arccos(433​−3​)​+32πn​,x=32π​−3arccos(433​−3​)​+32πn​
Combine all the solutionsx=3arccos(433​−3​)​+32πn​,x=32π​−3arccos(433​−3​)​+32πn​
Show solutions in decimal formx=30.81462…​+32πn​,x=32π​−30.81462…​+32πn​

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