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I'm not really sure what you mean by "the solution", but that equation cos = sec - sintan does simplify down to sin^2 + cos^2 = 1 which so happens to be an identity. I'm not sure if that's what you're looking for, but if it is, here are the steps in simplifying it. 1. Convert sec to 1/cos 2. Convert tan into sin/cos and multiply it by sin sintan = sin(sin/cos) = (sin^2)/cos You then have cos = 1/cos - (sin^2/cos) 3. Multiply everything by cos cos^2 = 1 - sin^2 4. And finally, send the sin^2 over to the left side by adding it (since it is being subracted on the right) You should see this sin^2 + cos^2 = 1 which is an identity.

Q: What is the solution to cos equals sec-sintan?

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cos x - 0.5 = 0 ⇒ cos x = 0.5 ⇒ x = 2nπ ± π/3

either cos OR tan-sin equals zero socos=0 at pi/2 and 3pi/2ortan=sin which is impossibleim not sure though

A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.

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

One solution. (cos x)2 - 2cos x = 3 Factor: (cos x - 3)(cos x + 1)= 0 cos x = {-1, 3} Solve: For cos x = -1, x = 180 deg No solution for cos x = 3

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cos x - 0.5 = 0 ⇒ cos x = 0.5 ⇒ x = 2nπ ± π/3

either cos OR tan-sin equals zero socos=0 at pi/2 and 3pi/2ortan=sin which is impossibleim not sure though

A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.

The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).

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

One solution. (cos x)2 - 2cos x = 3 Factor: (cos x - 3)(cos x + 1)= 0 cos x = {-1, 3} Solve: For cos x = -1, x = 180 deg No solution for cos x = 3

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.

y = 2 sin 3x y' = 2(sin 3x)'(3x)' y' = 2(cos 3x)(3) y' = 6 cos 3x

1. Anything divided by itself always equals 1.

If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²

cos x equals cos -x because cos is an even function. An even function f is such that f(x) = f(-x).