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Sin2x - radical 2 cosx equals 0?
Sin2x = radical 2
How do you solve sin2x plus 3 cos 2x equals 0?
sin2x + 3*cos2x = 0sin2x = -3*cos2xtan2x = -32x = arctan(-3)x = 0.5*arctan(-3) in the domain which should have been specified. As none has, the question has no answer.
What is the solution sets for sin3x plus sin2x plus sinx equals 0?
The solitions are in degrees. You may convert them to degrees should you wish. x= 0,90,120,180,240,270,360
Sin Squared x plus cos x equals 0 for x is a part of 1 to 360 degrees?
Replace sin2x with the equivalent (1 - cos2x). Simplify, and use the quadratic equation, to solve for cos x.Replace sin2x with the equivalent (1 - cos2x). Simplify, and use the quadratic equation, to solve for cos x.Replace sin2x with the equivalent (1 - cos2x). Simplify, and use the quadratic equation, to solve for cos x.Replace sin2x with the equivalent (1 - cos2x). Simplify, and use the quadratic equation, to solve for cos x.
sinx+sin2x=0?
SinX + Sin2X = 0 SinX + 2SinXCosX = 0 Factor SinX(1 - 2CosX) = 0 Hence SinX = 0 X = = 0 , 180, 360, .... & 1 - 2CosX = 0 2CosX = 1 Cos X = 1/2 = 0.5 X = 60, 240, 420, ....
Sin squared x pluss cosx equals 0?
It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.sin2x + cos x = 0(1 - cos2x) + cos x = 0-cos2x + cos x + 1 = 0cos2x - cos x - 1 = 0Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.sin2x + cos x = 0(1 - cos2x) + cos x = 0-cos2x + cos x + 1 = 0cos2x - cos x - 1 = 0Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.sin2x + cos x = 0(1 - cos2x) + cos x = 0-cos2x + cos x + 1 = 0cos2x - cos x - 1 = 0Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.sin2x + cos x = 0(1 - cos2x) + cos x = 0-cos2x + cos x + 1 = 0cos2x - cos x - 1 = 0Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.
What is the derivative of 1 divided by sinx?
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
Sin2x equals cos3x find x?
0. sin 2x = cos 3x 1. sin 2x = sin (pi/2 - 3x) [because cos u = sin (pi/2 - u)] 2. [...]
How do you solve sin squared theta plus cos theta equals sin theta plus cos squared theta?
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)
What is true 15 plus 0 equals 0 15-0 equals 0 15 percent 0 equals 0 0 percent 15 equals 0?
15 percent 0 equals 0; 0 percent 15 equals 0. Both the above are true
Ab equals 0 if and only if a equals 0 or b equals 0?
yes
Sin x - cos x 0?
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.
