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9(tan(2a)+2cos(a)=0 interval 0 less than or equal to a less than or equal to 2pi
Solve: sin x - 2sin x cos x = 0 for 0 is less than or equal to x is less than 2pi.
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: 2 cos2 x - 3 cos x + 1 = 0 for 0 is less than or equal to x is less than 2pi.
trig question2sin^2x+sinx-6=0how to solve for x when the restriction is x is greater than or equal to zero but less than 2pi
Solve for x in the equation 3sin^2x = cos^2x; 0
16 points) 18. Solve the equation of the interval [0, 2x) 2cos - Cos - 3 = 0 (6 points) 19. Use an identity to solve each equation on the interval [0, 2.1) sin 2x - 2006
16 points) 18. Solve the equation of the interval [0, 2x) 2cos - Cos - 3 = 0 (6 points) 19. Use an identity to solve each equation on the interval [0, 2.1) sin 2x - 2006
tan(x)^2=2tanxsinxsin(x)^2+sin(x)-1=03coscotx+7=5csc(x)2cos(x)^2-cos(x)=2-sec(x)I've tried to solve these problems so many time but can seem to figure out how to do it. I havethe ansers. i guess its just the alebraic manipulation the isthe problem .
If 2cos A-1=0 where 0<or = A<2pi then what is A equal to?
Solve the equation for x in the interval 0<x<2pi1/ 1+tan^2x = -cos x How would i do this? im thinking of maybe changing the 1+tan to sec^2x?