trig question
posted by Kate .
Compute inverse functions to four significant digits.
cos^2x=35cosx
cos^2x + 5cosx 3=0
now you have a quadratic, solve for cos x using the equation
cosx=(5 + sqrt (25 +12)/2
the back of the book says .9987+2kpi and .09987+2kpi
when i do it i don't get that answer.
I get it, that answer is in radians. The 2kpi is to keep rotation a number of times in the circle.
Thank you so much for helping me I have AC tomorrow and without your help i would have not been able to do that whole section.
Respond to this Question
Similar Questions

Trig
prove the identity (sinX)^6 +(cosX)^6= 1  3(sinX)^2 (cosX)^2 sinX^6= sinx^2 ^3 = (1cosX^2)^3 = (12CosX^2 + cos^4) (1cosX^2) then multiply that out 12CosX^2 + cos^4  cosX^2 + 2cos^4 cos^6 add that on the left to the cos^6, and … 
Trig
Compute inverse functions to four significant digits. cos^2x=35cosx rewrite it as.. cos^2x + 5cosx 3=0 now you have a quadratic, solve for cos x using the equation cosx=(5 + sqrt (25 +12)/2 
trig
it says solve for all real solutions, compute inverse functions to 4 sigfigs. cos^2x=35cosx Devon, what grade level are these questions from? 
Math  Solving for Trig Equations
Solve the following equation for 0 less than and/or equal to "x" less than and/or equal to 360  cos^2x  1 = sin^2x  Attempt: cos^2x  1  sin^2x = 0 cos^2x  1  (1  cos^2x) = 0 cos^2x  1  1 + cos^2x = 0 2cos^2x  2 = 0 (2cos^2x/2)= … 
Math  Solving Trig Equations
What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x)  1 cos(x) (+/)\sqrt{1  cos^2(x)} = 2cos^2(x)  1 cos^2(x)(1  cos^2(x)) 
Trigonometry
How do I solve these? 1) 2sinxcosxcosx=0 2) cos^2(x)0.5cosx=0 3) 6sin^2(x)5sinx+1=0 4) tan^2(x)+tanx12=0 
Trigonometry
Write equivalent equations in the form of inverse functions for a.)x=y+cos è b.)cosy=x^2 (can you show how you would solve) a.) x= y+ cos è cos è = xy theta = cos^1(xy) b.) cosy=x^2 cos(y) = x^2 y = Cos^1(x^2) 
Calc.
Differentiate. y= (cos x)^x u= cos x du= sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x1) * (sin x) =  x sin(x)cos^(x1)(x) (dy/dx)(dx/du)= … 
calculus
Differentiate. y= (cos x)^x u= cos x du= sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x1) * (sin x) =  x sin(x)cos^(x1)(x) (dy/dx)(dx/du)= … 
Trig/Precalc
So I have two questions that have been puzzling me for quite some time and would really appreciate any help with either of them! (a) There are four positive intergers a, b, c, and d such that 4cos(x)cos(2x)cos(4x)=cos(ax)+cos(bx)+cos(cx)+cos(dx) …