Sunday

February 1, 2015

February 1, 2015

Posted by **Kate** on Monday, April 16, 2007 at 8:27pm.

cos^2x=3-5cosx

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 dont 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.

**Answer this Question**

**Related Questions**

Trig - Compute inverse functions to four significant digits. cos^2x=3-5cosx ...

trig - it says solve for all real solutions, compute inverse functions to 4 ...

Math - Solving for Trig Equations - Solve the following equation for 0 less than...

Trig--check answer - Solve the equation of the interval (0, 2pi) cosx=sinx I ...

Trigonometry - Write equivalent equations in the form of inverse functions for a...

Trig - prove the identity (sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2 sinX^6= ...

Trigonometry - How do I solve these? 1) 2sinxcosx-cosx=0 2) cos^2(x)-0.5cosx=0 3...

calculus - Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x...

Calc. - Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln...

Math - Solving Trig Equations - What am I doing wrong? Equation: sin2x = 2cos2x ...