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Find the value of x in the interval 0 is less than or equal to x and 360 is greater than or equal to x which satisfies the equation cos x - 2 cos x sin x
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 = 0cos^2x - 1 - (1 - cos^2x) = 0cos^2x - 1 - 1 + cos^2x = 02cos^2x - 2 = 0(2cos^2x/2)= (-2/2)cos^2x = -1cosx = square root of -1And I can't do anything with this now...what am I doing wrong?--Textbook answers:0, 180, 360--
Solve each equation for o is less than and/or equal to theta is less than and/or equal to 360--sin^2x = 1 = cos^2x--Work:cos^2x - cos^2x = 00 = 0--Textbook Answers:90 and 270--Btw, how would you isolate for cos^2x = 0? Would it be...x = cos^-1 square root of _____ ?
Solve: 2 sin x - 1 = 0 for 0 degrees is less than or equal to x is less than 360 degrees.
Solve to the nearest minute: x is greater than or equal to 0 and less than 360.3sec^2 x-8tan-6=cotx
9(tan(2a)+2cos(a)=0 interval 0 less than or equal to a less than or equal to 2piSolve each equation...trig equations for linear sin and cos
Solve: sin x - 2sin x cos x = 0 for 0 is less than or equal to x is less than 2pi.
solve for the exact value of x where 0 greater than or equal to X and less than/equal to 2 pi. cos^2 x-1=0 I did the factoring and FOILing and The two answers I got were: cosx=-1 cosx=1 x=pi x=0,2pi According to my instructor cosx=1's answers are not completly correct. I dont see what the problem is since cos(0) and cos(2pi) both =1. Am I missing something here? Thank you for your help.
Solve & plot a graph for the following: F(X)=cos(X)-cos to second power X for the interval negative pi greater than or equal to X less than or equal to positive pi
For 2cos^2X + cosX - 1= 0 find X 0<X<360 <=greater than or equal to.
Solve the equation if 0 is less than or equal to x and x is less than 2 pie 2sin^2x+3 cos x -3 0