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Q: 2 cos x plus 1 cos x plus -1 equals 0?

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No. sin(0) = 0 So cos(0)*sin(0) = 0 so the left hand side = 1

a = 0, b = 0.

cos x - 1 = 0 cos(x) = 1 x = 0 +/- k*pi radians where k = 1,2,3,...

sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.

No, but cos(-x) = cos(x), because the cosine function is an even function.

1

Yes because if 1+0=1 than 0 plus b equals b

1 plus 1 plus 1 plus 1 equals 1 times 4. 1 times 4 equals 4. 4 minus 4 equals 0. 0

[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,

It equals 7

It equals 6

sqrt(3sin(x)=cos(x)=0 // Square both sides3sin(x) + cos(x) = 0 // subtract cos(x) from both sides3sin(x) = -cos(x) // rearrangesin(x)/cos(x) = -1/3 //sin(x)/cos(x) = tan(x)tan(x) = -1/3x = tan^-1(-1/3) == -18,43484882 // tan^-1(inverse tan)

If it's 1 +-1 =0

Secant x= 1/cosx So if cos x=1 ,we know that x=0 degrees ( or radians), so secant x is 1/cos (0)=1/1=1

2

cos2(theta) = 1 so cos(theta) = Â±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0

No, (sinx)^2 + (cosx)^2=1 is though

x = 0

The fisrt thing to note is that there is no variable in the question and so there cannot be a solution set. The only possibilities are the statement is true, false or indeterminate. Now, 2 cos 0 - 1 = 0 is equivalent to 2*1 - 1 = 0 [since cos(0) = 1] or 2 - 1 = 0 which is false.

0 because X= -1

because when you add 1 to 0 it is 2

Not in normal arithmetic.

1

You cannot prove it because it is not true! cos(0) = 1 cos(2*pi) = 1 cos(4*pi) = 1 ... cos(2*k*pi) = 1 for all integers k or, if you still work in degrees, cos(0) = 1 cos(360) = 1 cos(720) = 1 ... cos(k*360) = 1 for all integers k

Anything multiplied by 0 is 0, so all the ones before the 0 can be ignored. It is now 0+1 which equals 1.