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Popular Trigonometry >

sec(x)+sin^2(x)+cos^2(x)=tan^2(x)

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Solution

sec(x)+sin2(x)+cos2(x)=tan2(x)

Solution

x=π+2πn,x=3π​+2πn,x=35π​+2πn
+1
Degrees
x=180∘+360∘n,x=60∘+360∘n,x=300∘+360∘n
Solution steps
sec(x)+sin2(x)+cos2(x)=tan2(x)
Subtract tan2(x) from both sidessec(x)+sin2(x)+cos2(x)−tan2(x)=0
Express with sin, cos
cos2(x)+sec(x)+sin2(x)−tan2(x)
Use the basic trigonometric identity: sec(x)=cos(x)1​=cos2(x)+cos(x)1​+sin2(x)−tan2(x)
Use the basic trigonometric identity: tan(x)=cos(x)sin(x)​=cos2(x)+cos(x)1​+sin2(x)−(cos(x)sin(x)​)2
Simplify cos2(x)+cos(x)1​+sin2(x)−(cos(x)sin(x)​)2:cos2(x)cos4(x)+cos(x)+sin2(x)cos2(x)−sin2(x)​
cos2(x)+cos(x)1​+sin2(x)−(cos(x)sin(x)​)2
Apply exponent rule: (ba​)c=bcac​=cos2(x)+cos(x)1​+sin2(x)−cos2(x)sin2(x)​
Convert element to fraction: cos2(x)=1cos2(x)​,sin2(x)=1sin2(x)​=1cos2(x)​+cos(x)1​+1sin2(x)​−cos2(x)sin2(x)​
Least Common Multiplier of 1,cos(x),1,cos2(x):cos2(x)
1,cos(x),1,cos2(x)
Lowest Common Multiplier (LCM)
Least Common Multiplier of 1,1:1
1,1
Least Common Multiplier (LCM)
Prime factorization of 1
Prime factorization of 1
Multiply each factor the greatest number of times it occurs in either 1 or 1=1
Multiply the numbers: 1=1=1
Compute an expression comprised of factors that appear in at least one of the factored expressions=cos2(x)
Adjust Fractions based on the LCM
Multiply each numerator by the same amount needed to multiply its
corresponding denominator to turn it into the LCM cos2(x)
For 1cos2(x)​:multiply the denominator and numerator by cos2(x)1cos2(x)​=1⋅cos2(x)cos2(x)cos2(x)​=cos2(x)cos4(x)​
For cos(x)1​:multiply the denominator and numerator by cos(x)cos(x)1​=cos(x)cos(x)1⋅cos(x)​=cos2(x)cos(x)​
For 1sin2(x)​:multiply the denominator and numerator by cos2(x)1sin2(x)​=1⋅cos2(x)sin2(x)cos2(x)​=cos2(x)sin2(x)cos2(x)​
=cos2(x)cos4(x)​+cos2(x)cos(x)​+cos2(x)sin2(x)cos2(x)​−cos2(x)sin2(x)​
Since the denominators are equal, combine the fractions: ca​±cb​=ca±b​=cos2(x)cos4(x)+cos(x)+sin2(x)cos2(x)−sin2(x)​
=cos2(x)cos4(x)+cos(x)+sin2(x)cos2(x)−sin2(x)​
cos2(x)cos(x)+cos4(x)−sin2(x)+cos2(x)sin2(x)​=0
g(x)f(x)​=0⇒f(x)=0cos(x)+cos4(x)−sin2(x)+cos2(x)sin2(x)=0
Factor cos(x)+cos4(x)−sin2(x)+cos2(x)sin2(x):(1+cos(x))(cos(x)(cos2(x)+1−cos(x))+sin2(x)(−1+cos(x)))
cos(x)+cos4(x)−sin2(x)+cos2(x)sin2(x)
Factor cos(x)+cos4(x):cos(x)(cos(x)+1)(cos2(x)−cos(x)+1)
cos(x)+cos4(x)
Apply exponent rule: ab+c=abaccos4(x)=cos(x)cos3(x)=cos(x)+cos(x)cos3(x)
Factor out common term cos(x)=cos(x)(1+cos3(x))
Factor cos3(x)+1:(cos(x)+1)(cos2(x)−cos(x)+1)
1+cos3(x)
Rewrite 1 as 13=cos3(x)+13
Apply Sum of Cubes Formula: x3+y3=(x+y)(x2−xy+y2)cos3(x)+13=(cos(x)+1)(cos2(x)−cos(x)+1)=(cos(x)+1)(cos2(x)−cos(x)+1)
=cos(x)(cos(x)+1)(cos2(x)−cos(x)+1)
Factor −sin2(x)+cos2(x)sin2(x):sin2(x)(cos(x)+1)(cos(x)−1)
−sin2(x)+cos2(x)sin2(x)
Factor out common term sin2(x)=sin2(x)(−1+cos2(x))
Factor cos2(x)−1:(cos(x)+1)(cos(x)−1)
−1+cos2(x)
Rewrite 1 as 12=cos2(x)−12
Apply Difference of Two Squares Formula: x2−y2=(x+y)(x−y)cos2(x)−12=(cos(x)+1)(cos(x)−1)=(cos(x)+1)(cos(x)−1)
=sin2(x)(cos(x)+1)(cos(x)−1)
=cos(x)(cos(x)+1)(cos2(x)−cos(x)+1)+sin2(x)(cos(x)+1)(cos(x)−1)
Factor out common term (1+cos(x))=(1+cos(x))(cos(x)(cos2(x)+1−cos(x))+sin2(x)(−1+cos(x)))
(1+cos(x))(cos(x)(cos2(x)+1−cos(x))+sin2(x)(−1+cos(x)))=0
Solving each part separately1+cos(x)=0orcos(x)(cos2(x)+1−cos(x))+sin2(x)(−1+cos(x))=0
1+cos(x)=0:x=π+2πn
1+cos(x)=0
Move 1to the right side
1+cos(x)=0
Subtract 1 from both sides1+cos(x)−1=0−1
Simplifycos(x)=−1
cos(x)=−1
General solutions for cos(x)=−1
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
x=π+2πn
x=π+2πn
cos(x)(cos2(x)+1−cos(x))+sin2(x)(−1+cos(x))=0:x=3π​+2πn,x=35π​+2πn
cos(x)(cos2(x)+1−cos(x))+sin2(x)(−1+cos(x))=0
Rewrite using trig identities
(−1+cos(x))sin2(x)+(1−cos(x)+cos2(x))cos(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1cos2(x)=1−sin2(x)=(−1+cos(x))sin2(x)+(1−cos(x)+1−sin2(x))cos(x)
Simplify (−1+cos(x))sin2(x)+(1−cos(x)+1−sin2(x))cos(x):−sin2(x)−cos2(x)+2cos(x)
(−1+cos(x))sin2(x)+(1−cos(x)+1−sin2(x))cos(x)
Simplify 1−cos(x)+1−sin2(x):−sin2(x)−cos(x)+2
1−cos(x)+1−sin2(x)
Group like terms=−cos(x)−sin2(x)+1+1
Add the numbers: 1+1=2=−sin2(x)−cos(x)+2
=sin2(x)(cos(x)−1)+cos(x)(−sin2(x)−cos(x)+2)
=sin2(x)(−1+cos(x))+cos(x)(−sin2(x)−cos(x)+2)
Expand sin2(x)(−1+cos(x)):−sin2(x)+sin2(x)cos(x)
sin2(x)(−1+cos(x))
Apply the distributive law: a(b+c)=ab+aca=sin2(x),b=−1,c=cos(x)=sin2(x)(−1)+sin2(x)cos(x)
Apply minus-plus rules+(−a)=−a=−1⋅sin2(x)+sin2(x)cos(x)
Multiply: 1⋅sin2(x)=sin2(x)=−sin2(x)+sin2(x)cos(x)
=−sin2(x)+sin2(x)cos(x)+(−sin2(x)−cos(x)+2)cos(x)
Expand cos(x)(−sin2(x)−cos(x)+2):−sin2(x)cos(x)−cos2(x)+2cos(x)
cos(x)(−sin2(x)−cos(x)+2)
Distribute parentheses=cos(x)(−sin2(x))+cos(x)(−cos(x))+cos(x)⋅2
Apply minus-plus rules+(−a)=−a=−sin2(x)cos(x)−cos(x)cos(x)+2cos(x)
cos(x)cos(x)=cos2(x)
cos(x)cos(x)
Apply exponent rule: ab⋅ac=ab+ccos(x)cos(x)=cos1+1(x)=cos1+1(x)
Add the numbers: 1+1=2=cos2(x)
=−sin2(x)cos(x)−cos2(x)+2cos(x)
=−sin2(x)+sin2(x)cos(x)−sin2(x)cos(x)−cos2(x)+2cos(x)
Add similar elements: sin2(x)cos(x)−sin2(x)cos(x)=0=−sin2(x)−cos2(x)+2cos(x)
=−sin2(x)−cos2(x)+2cos(x)
Use the Pythagorean identity: cos2(x)+sin2(x)=1−cos2(x)−sin2(x)=−1=2cos(x)−1
2cos(x)−1=0
Move 1to the right side
2cos(x)−1=0
Add 1 to both sides2cos(x)−1+1=0+1
Simplify2cos(x)=1
2cos(x)=1
Divide both sides by 2
2cos(x)=1
Divide both sides by 222cos(x)​=21​
Simplifycos(x)=21​
cos(x)=21​
General solutions for cos(x)=21​
cos(x) periodicity table with 2πn cycle:
x06π​4π​3π​2π​32π​43π​65π​​cos(x)123​​22​​21​0−21​−22​​−23​​​xπ67π​45π​34π​23π​35π​47π​611π​​cos(x)−1−23​​−22​​−21​021​22​​23​​​​
x=3π​+2πn,x=35π​+2πn
x=3π​+2πn,x=35π​+2πn
Combine all the solutionsx=π+2πn,x=3π​+2πn,x=35π​+2πn

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Frequently Asked Questions (FAQ)

  • What is the general solution for sec(x)+sin^2(x)+cos^2(x)=tan^2(x) ?

    The general solution for sec(x)+sin^2(x)+cos^2(x)=tan^2(x) is x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin
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