The statement of the problem is equivalent to sin x = - cos x. This is true for x = 135 degrees and x = -45 degrees, and also for (135 + 180n) degrees, where n is any integer.
No, but cos(-x) = cos(x), because the cosine function is an even function.
Sin 15 + cos 105 = -1.9045
The question contains an expression but not an equation. An expression cannot be solved.
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.
3cos
Better formatting is cos(2x+20)=-0.5
No, but cos(-x) = cos(x), because the cosine function is an even function.
Sin 15 + cos 105 = -1.9045
For the product to be zero, any of the factors must be zero, so you solve, separately, the two equations: sin x = 0 and: cos x = 0 Like many trigonometric equations, this will have an infinity of solutions, since sine and cosine are periodic functions.
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.
1
cos x - 1 = 0 cos(x) = 1 x = 0 +/- k*pi radians where k = 1,2,3,...
The question contains an expression but not an equation. An expression cannot be solved.
To solve the equation 2cos(x) + cos(x) - 1 = 0, we first combine like terms to get 3cos(x) - 1 = 0. Then, we isolate the cosine term by adding 1 to both sides to get 3cos(x) = 1. Finally, we divide by 3 to solve for cos(x), which gives cos(x) = 1/3. Therefore, x = arccos(1/3) or approximately 70.53 degrees.
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.
3cos
[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,