2sinxcosx-cosx=0 Factored : cosx(2sinx-1)=0 2 solutions: cosx=0 or sinx=.5 For cosx=0, x=90 or 270 degrees For sinx=.5, x=30 degrees x = {30, 90, 270}
cosx + sinx = 0 when sinx = -cosx. By dividing both sides by cosx you get: sinx/cosx = -1 tanx = -1 The values where tanx = -1 are 3pi/4, 7pi/4, etc. Those are equivalent to 135 degrees, 315 degrees, etc.
You can move it up or down by adding a constant, call it c. Let c>0 Y=radical(x)+c move it up c and y= radical(x)-c moves it down c. You can move it to the right by subtracting c inside the radical sign. Let c>0 y=radical (x-c) moves it to the right c units. y=radical (x+c) moves it to the left c units.
It is a quadratic equation and can be rearranged in the form of:- x2-x-6 = 0 (x+2)(x-3) = 0 Solutions: x = -2 and x = 3
Undefined: You cannot divide by zero
Sinx = 0 CosX= -1/2
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
0
tanx=2cscx sinx/cosx=2/sinx sin2x/cosx=2 sin2x=2cosx 1-cos2x=2cosx 0=cos2x+2cosx-1 Quadratic formula: cosx=(-2±√(2^2+4))/2 cosx=(-2±√8)/2 cosx=(-2±2√2)/2 cosx=-1±√2 cosx=approximately -2.41 or approximately 0.41. Since the range of the cosine function is [-1,1], only approx. 0.41 works. So: cosx= approx. 0.41 Need calculator now (I went as far as I could without one!) x=approx 1.148
2sinxcosx-cosx=0 Factored : cosx(2sinx-1)=0 2 solutions: cosx=0 or sinx=.5 For cosx=0, x=90 or 270 degrees For sinx=.5, x=30 degrees x = {30, 90, 270}
x = 3pi/4
For simplicity's sake, X represent theta. This is the original problem: sin2x+ cosX = cos2X + sinX This handy-dandy property is key for all you trig fanatics: sin2x+ cos2x = 1 With this basic property, you can figure out that sin2 x=1-cos2x and cos2x= 1-sin2x So we can change the original problem to: 1-cos2x+cosx = 1-sin2X + sinX -cos2x + cosx =-sin2x + sinX Basic logic tells you that one of two things are happening. sin2x is equal to sinx AND cos2x is equal to cosx. The only two numbers that are the same squared as they are to the first power are 1 and 0. X could equal 0, which has a cosine of 1 and a sine of 0, or it could equal pi/2, which has a cosine of 0 and a sine of 1. The other possibility whatever x (or theta) is, it's sine is equal to its cosine. This happens twice on the unit circle, once at pi/4 and once at 5pi/4. If you're solving for all possible values for x and not just a set range on the unit circle, then the final solution is: x=0+2pin x=pi/2+2pin x= pi/4 +2pin x=5pi/4+2pin (note that n is a variable)
2cos2x - cosx -1 = 0 Factor: (2cosx + 1)(cosx - 1) = 0 cosx = {-.5, 1} x = {...0, 120, 240, 360,...} degrees
sinx-cosx=0 --> move cosx to opposite side sinx=cosx --> square both sides sin2x=cos2x --> use pythagorean identities for (cos2x=1-sin2x) sin2x=1-sin2x --> add sin2x to both sides of equation 2sin2x=1 --> divide both sides by 2 sin2x=1/2 --> take the square root of both sides sinx= +/- (square root of 2)/2 or .7071 If giving answers in radians --> answer appears in all four quadrants, so answer would be (pi/4 + piN/2). Other answers would be (3pi/4 + piN/2), (5pi/4 + piN/2), and (7pi/4 + piN/2). Check for extraneous solutions: The answers in the first and third quadrant are extraneous. Therefore, your answer is (3pi/4 + piN), because every pi, an answer occurs. In one trip around the quadrants, both 3pi/4 and 7pi/4 are answers.
cosx + sinx = 0 when sinx = -cosx. By dividing both sides by cosx you get: sinx/cosx = -1 tanx = -1 The values where tanx = -1 are 3pi/4, 7pi/4, etc. Those are equivalent to 135 degrees, 315 degrees, etc.
It is sqrt(0), which equals 0.
X=60 how did you get that? could you show all the steps?