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Popular Trigonometry >

solvefor x,cos^2(x)*z^2-2cos^2(x)*z+1=0

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Solution

solvefor

Solution

NoSolutionforx∈R
Solution steps
cos2(x)z2−2cos2(x)z+1=0
Solve by substitution
cos2(x)z2−2cos2(x)z+1=0
Let: cos(x)=uu2z2−2u2z+1=0
u2z2−2u2z+1=0:u=iz2−2z1​​,u=−iz2−2z1​​;z=0,z=2
u2z2−2u2z+1=0
Move 1to the right side
u2z2−2u2z+1=0
Subtract 1 from both sidesu2z2−2u2z+1−1=0−1
Simplifyu2z2−2u2z=−1
u2z2−2u2z=−1
Factor u2z2−2u2z:zu2(z−2)
u2z2−2u2z
Apply exponent rule: ab+c=abacz2=zz=u2zz−2u2z
Factor out common term u2z=u2z(z−2)
zu2(z−2)=−1
Divide both sides by z(z−2);z=0,z=2
zu2(z−2)=−1
Divide both sides by z(z−2);z=0,z=2z(z−2)zu2(z−2)​=z(z−2)−1​;z=0,z=2
Simplifyu2=−z(z−2)1​;z=0,z=2
u2=−z(z−2)1​;z=0,z=2
For x2=f(a) the solutions are x=f(a)​,−f(a)​
u=−z(z−2)1​​,u=−−z(z−2)1​​;z=0,z=2
Simplify −z(z−2)1​​:iz2−2z1​​
−z(z−2)1​​
Apply radical rule: −a​=−1​a​, assuming a≥0−z(z−2)1​​=−1​z(z−2)1​​=−1​z(z−2)1​​
Apply imaginary number rule: −1​=i=iz(z−2)1​​
Rewrite iz(z−2)1​​ in standard complex form: z2−2z1​​i
iz(z−2)1​​
z(z−2)1​​=z2−2z1​​
z(z−2)1​​
z(z−2)1​=z2−2z1​
z(z−2)1​
Expand z(z−2):z2−2z
z(z−2)
Apply the distributive law: a(b−c)=ab−aca=z,b=z,c=2=zz−z⋅2
=zz−2z
zz=z2
zz
Apply exponent rule: ab⋅ac=ab+czz=z1+1=z1+1
Add the numbers: 1+1=2=z2
=z2−2z
=z2−2z1​
=z2−2z1​​
=iz2−2z1​​
=z2−2z1​​i
Simplify −−z(z−2)1​​:−iz2−2z1​​
−−z(z−2)1​​
Simplify −z(z−2)1​​:iz(z−2)1​​
−z(z−2)1​​
Apply radical rule: −a​=−1​a​, assuming a≥0−z(z−2)1​​=−1​z(z−2)1​​=−1​z(z−2)1​​
Apply imaginary number rule: −1​=i=iz(z−2)1​​
=−iz(z−2)1​​
Rewrite −iz(z−2)1​​ in standard complex form: −z2−2z1​​i
−iz(z−2)1​​
z(z−2)1​​=z2−2z1​​
z(z−2)1​​
z(z−2)1​=z2−2z1​
z(z−2)1​
Expand z(z−2):z2−2z
z(z−2)
Apply the distributive law: a(b−c)=ab−aca=z,b=z,c=2=zz−z⋅2
=zz−2z
zz=z2
zz
Apply exponent rule: ab⋅ac=ab+czz=z1+1=z1+1
Add the numbers: 1+1=2=z2
=z2−2z
=z2−2z1​
=z2−2z1​​
=−iz2−2z1​​
=−z2−2z1​​i
u=iz2−2z1​​,u=−iz2−2z1​​;z=0,z=2
Substitute back u=cos(x)cos(x)=iz2−2z1​​,cos(x)=−iz2−2z1​​;z=0,z=2
cos(x)=iz2−2z1​​,cos(x)=−iz2−2z1​​;z=0,z=2
cos(x)=iz2−2z1​​:No Solution
cos(x)=iz2−2z1​​
NoSolution
cos(x)=−iz2−2z1​​:No Solution
cos(x)=−iz2−2z1​​
NoSolution
Combine all the solutionsNoSolutionforx∈R

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Frequently Asked Questions (FAQ)

  • What is the general solution for solvefor x,cos^2(x)*z^2-2cos^2(x)*z+1=0 ?

    The general solution for solvefor x,cos^2(x)*z^2-2cos^2(x)*z+1=0 is No Solution for x\in\mathbb{R}
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