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juantheron
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Number of real solution of the equation $\sin (\sin (\sin x)) = \cos (\cos (\cos x))$
where $ 0 \leq x\leq 2\pi$
where $ 0 \leq x\leq 2\pi$
jacks said:Number of real solution <--- 2
of the equation $\sin (\sin (\sin x)) = \cos (\cos (\cos x))$
where $ 0 \leq x\leq 2\pi$
The number of real solutions for a trigonometric equation can be determined by looking at the coefficients and constants in the equation. If the equation has only one trigonometric function with a coefficient of 1, then it will have exactly two real solutions. If the equation has multiple trigonometric functions or a coefficient other than 1, then the number of real solutions cannot be determined without solving the equation.
Yes, a trigonometric equation can have no real solutions. This occurs when the equation has no solutions that satisfy both the trigonometric function and the domain restrictions, such as when the equation involves a tangent or cotangent function and the angle is restricted to certain values.
Real solutions for a trigonometric equation are values that satisfy both the equation and the domain restrictions, and can be represented on the real number line. Complex solutions, on the other hand, are values that satisfy the equation but do not satisfy the domain restrictions, and therefore cannot be represented on the real number line.
Yes, a trigonometric equation can have an infinite number of solutions. This occurs when the equation has a periodic function, such as sine or cosine, and the angle can take on an infinite number of values that satisfy the equation.
To solve a trigonometric equation with multiple solutions, you can use the unit circle or graphing techniques to find the solutions within a specified interval. You can also use the general solutions, which involve adding or subtracting multiples of the period of the trigonometric function to the principal solution.