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Popular Trigonometry >

sin^4(x)+cos^4(x)= 7/9

  • Pre Algebra
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Solution

sin4(x)+cos4(x)=97​

Solution

x=0.36486…+2πn,x=2π−0.36486…+2πn,x=2.77672…+2πn,x=−2.77672…+2πn,x=1.20593…+2πn,x=2π−1.20593…+2πn,x=1.93566…+2πn,x=−1.93566…+2πn
+1
Degrees
x=20.90515…∘+360∘n,x=339.09484…∘+360∘n,x=159.09484…∘+360∘n,x=−159.09484…∘+360∘n,x=69.09484…∘+360∘n,x=290.90515…∘+360∘n,x=110.90515…∘+360∘n,x=−110.90515…∘+360∘n
Solution steps
sin4(x)+cos4(x)=97​
Subtract 97​ from both sidessin4(x)+cos4(x)−97​=0
Simplify sin4(x)+cos4(x)−97​:99sin4(x)+9cos4(x)−7​
sin4(x)+cos4(x)−97​
Convert element to fraction: sin4(x)=9sin4(x)9​,cos4(x)=9cos4(x)9​=9sin4(x)⋅9​+9cos4(x)⋅9​−97​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=9sin4(x)⋅9+cos4(x)⋅9−7​
99sin4(x)+9cos4(x)−7​=0
g(x)f(x)​=0⇒f(x)=09sin4(x)+9cos4(x)−7=0
Apply exponent rule: ab=a2ab−2−7+9cos4(x)+9sin2(x)sin2(x)=0
Rewrite using trig identities
−7+9cos4(x)+9sin2(x)sin2(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1sin2(x)=1−cos2(x)=−7+9cos4(x)+9(1−cos2(x))(1−cos2(x))
Simplify −7+9cos4(x)+9(1−cos2(x))(1−cos2(x)):18cos4(x)−18cos2(x)+2
−7+9cos4(x)+9(1−cos2(x))(1−cos2(x))
9(1−cos2(x))(1−cos2(x))=9(1−cos2(x))2
9(1−cos2(x))(1−cos2(x))
Apply exponent rule: ab⋅ac=ab+c(1−cos2(x))(1−cos2(x))=(1−cos2(x))1+1=9(1−cos2(x))1+1
Add the numbers: 1+1=2=9(1−cos2(x))2
=−7+9cos4(x)+9(−cos2(x)+1)2
(1−cos2(x))2:1−2cos2(x)+cos4(x)
Apply Perfect Square Formula: (a−b)2=a2−2ab+b2a=1,b=cos2(x)
=12−2⋅1⋅cos2(x)+(cos2(x))2
Simplify 12−2⋅1⋅cos2(x)+(cos2(x))2:1−2cos2(x)+cos4(x)
12−2⋅1⋅cos2(x)+(cos2(x))2
Apply rule 1a=112=1=1−2⋅1⋅cos2(x)+(cos2(x))2
2⋅1⋅cos2(x)=2cos2(x)
2⋅1⋅cos2(x)
Multiply the numbers: 2⋅1=2=2cos2(x)
(cos2(x))2=cos4(x)
(cos2(x))2
Apply exponent rule: (ab)c=abc=cos2⋅2(x)
Multiply the numbers: 2⋅2=4=cos4(x)
=1−2cos2(x)+cos4(x)
=1−2cos2(x)+cos4(x)
=−7+9cos4(x)+9(1−2cos2(x)+cos4(x))
Expand 9(1−2cos2(x)+cos4(x)):9−18cos2(x)+9cos4(x)
9(1−2cos2(x)+cos4(x))
Distribute parentheses=9⋅1+9(−2cos2(x))+9cos4(x)
Apply minus-plus rules+(−a)=−a=9⋅1−9⋅2cos2(x)+9cos4(x)
Simplify 9⋅1−9⋅2cos2(x)+9cos4(x):9−18cos2(x)+9cos4(x)
9⋅1−9⋅2cos2(x)+9cos4(x)
Multiply the numbers: 9⋅1=9=9−9⋅2cos2(x)+9cos4(x)
Multiply the numbers: 9⋅2=18=9−18cos2(x)+9cos4(x)
=9−18cos2(x)+9cos4(x)
=−7+9cos4(x)+9−18cos2(x)+9cos4(x)
Simplify −7+9cos4(x)+9−18cos2(x)+9cos4(x):18cos4(x)−18cos2(x)+2
−7+9cos4(x)+9−18cos2(x)+9cos4(x)
Group like terms=9cos4(x)−18cos2(x)+9cos4(x)−7+9
Add similar elements: 9cos4(x)+9cos4(x)=18cos4(x)=18cos4(x)−18cos2(x)−7+9
Add/Subtract the numbers: −7+9=2=18cos4(x)−18cos2(x)+2
=18cos4(x)−18cos2(x)+2
=18cos4(x)−18cos2(x)+2
2−18cos2(x)+18cos4(x)=0
Solve by substitution
2−18cos2(x)+18cos4(x)=0
Let: cos(x)=u2−18u2+18u4=0
2−18u2+18u4=0:u=63+5​​​,u=−63+5​​​,u=63−5​​​,u=−63−5​​​
2−18u2+18u4=0
Write in the standard form an​xn+…+a1​x+a0​=018u4−18u2+2=0
Rewrite the equation with v=u2 and v2=u418v2−18v+2=0
Solve 18v2−18v+2=0:v=63+5​​,v=63−5​​
18v2−18v+2=0
Solve with the quadratic formula
18v2−18v+2=0
Quadratic Equation Formula:
For a=18,b=−18,c=2v1,2​=2⋅18−(−18)±(−18)2−4⋅18⋅2​​
v1,2​=2⋅18−(−18)±(−18)2−4⋅18⋅2​​
(−18)2−4⋅18⋅2​=65​
(−18)2−4⋅18⋅2​
Apply exponent rule: (−a)n=an,if n is even(−18)2=182=182−4⋅18⋅2​
Multiply the numbers: 4⋅18⋅2=144=182−144​
182=324=324−144​
Subtract the numbers: 324−144=180=180​
Prime factorization of 180:22⋅32⋅5
180
180divides by 2180=90⋅2=2⋅90
90divides by 290=45⋅2=2⋅2⋅45
45divides by 345=15⋅3=2⋅2⋅3⋅15
15divides by 315=5⋅3=2⋅2⋅3⋅3⋅5
2,3,5 are all prime numbers, therefore no further factorization is possible=2⋅2⋅3⋅3⋅5
=22⋅32⋅5
=22⋅32⋅5​
Apply radical rule: nab​=na​nb​=5​22​32​
Apply radical rule: nan​=a22​=2=25​32​
Apply radical rule: nan​=a32​=3=2⋅35​
Refine=65​
v1,2​=2⋅18−(−18)±65​​
Separate the solutionsv1​=2⋅18−(−18)+65​​,v2​=2⋅18−(−18)−65​​
v=2⋅18−(−18)+65​​:63+5​​
2⋅18−(−18)+65​​
Apply rule −(−a)=a=2⋅1818+65​​
Multiply the numbers: 2⋅18=36=3618+65​​
Factor 18+65​:6(3+5​)
18+65​
Rewrite as=6⋅3+65​
Factor out common term 6=6(3+5​)
=366(3+5​)​
Cancel the common factor: 6=63+5​​
v=2⋅18−(−18)−65​​:63−5​​
2⋅18−(−18)−65​​
Apply rule −(−a)=a=2⋅1818−65​​
Multiply the numbers: 2⋅18=36=3618−65​​
Factor 18−65​:6(3−5​)
18−65​
Rewrite as=6⋅3−65​
Factor out common term 6=6(3−5​)
=366(3−5​)​
Cancel the common factor: 6=63−5​​
The solutions to the quadratic equation are:v=63+5​​,v=63−5​​
v=63+5​​,v=63−5​​
Substitute back v=u2,solve for u
Solve u2=63+5​​:u=63+5​​​,u=−63+5​​​
u2=63+5​​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=63+5​​​,u=−63+5​​​
Solve u2=63−5​​:u=63−5​​​,u=−63−5​​​
u2=63−5​​
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=63−5​​​,u=−63−5​​​
The solutions are
u=63+5​​​,u=−63+5​​​,u=63−5​​​,u=−63−5​​​
Substitute back u=cos(x)cos(x)=63+5​​​,cos(x)=−63+5​​​,cos(x)=63−5​​​,cos(x)=−63−5​​​
cos(x)=63+5​​​,cos(x)=−63+5​​​,cos(x)=63−5​​​,cos(x)=−63−5​​​
cos(x)=63+5​​​:x=arccos​63+5​​​​+2πn,x=2π−arccos​63+5​​​​+2πn
cos(x)=63+5​​​
Apply trig inverse properties
cos(x)=63+5​​​
General solutions for cos(x)=63+5​​​cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos​63+5​​​​+2πn,x=2π−arccos​63+5​​​​+2πn
x=arccos​63+5​​​​+2πn,x=2π−arccos​63+5​​​​+2πn
cos(x)=−63+5​​​:x=arccos​−63+5​​​​+2πn,x=−arccos​−63+5​​​​+2πn
cos(x)=−63+5​​​
Apply trig inverse properties
cos(x)=−63+5​​​
General solutions for cos(x)=−63+5​​​cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos​−63+5​​​​+2πn,x=−arccos​−63+5​​​​+2πn
x=arccos​−63+5​​​​+2πn,x=−arccos​−63+5​​​​+2πn
cos(x)=63−5​​​:x=arccos​63−5​​​​+2πn,x=2π−arccos​63−5​​​​+2πn
cos(x)=63−5​​​
Apply trig inverse properties
cos(x)=63−5​​​
General solutions for cos(x)=63−5​​​cos(x)=a⇒x=arccos(a)+2πn,x=2π−arccos(a)+2πnx=arccos​63−5​​​​+2πn,x=2π−arccos​63−5​​​​+2πn
x=arccos​63−5​​​​+2πn,x=2π−arccos​63−5​​​​+2πn
cos(x)=−63−5​​​:x=arccos​−63−5​​​​+2πn,x=−arccos​−63−5​​​​+2πn
cos(x)=−63−5​​​
Apply trig inverse properties
cos(x)=−63−5​​​
General solutions for cos(x)=−63−5​​​cos(x)=−a⇒x=arccos(−a)+2πn,x=−arccos(−a)+2πnx=arccos​−63−5​​​​+2πn,x=−arccos​−63−5​​​​+2πn
x=arccos​−63−5​​​​+2πn,x=−arccos​−63−5​​​​+2πn
Combine all the solutionsx=arccos​63+5​​​​+2πn,x=2π−arccos​63+5​​​​+2πn,x=arccos​−63+5​​​​+2πn,x=−arccos​−63+5​​​​+2πn,x=arccos​63−5​​​​+2πn,x=2π−arccos​63−5​​​​+2πn,x=arccos​−63−5​​​​+2πn,x=−arccos​−63−5​​​​+2πn
Show solutions in decimal formx=0.36486…+2πn,x=2π−0.36486…+2πn,x=2.77672…+2πn,x=−2.77672…+2πn,x=1.20593…+2πn,x=2π−1.20593…+2πn,x=1.93566…+2πn,x=−1.93566…+2πn

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