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Calculte the number of solutions for  2\sin^2 x+3\cos 2x-3=0 where x\epsilon [0,2\pi]

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

0


Answers (1)

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Trigonometric Equations

Trigonometric equations are, as the name implies, equations that involve trigonometric functions.

Solution of Trigonometric Equation

The value of an unknown angle which satisfies the given trigonometric equation is called a solution or root of the equation. For example, 2sin? = 1, clearly ? = 300 satisfies the equation; therefore, 30is a solution of the equation. Now trigonometric equation ususally has infinite solutions due to periodic nature of trigonometric functions. So this equation also has (360+30)o,(720+30)o,(-360+30)o and so on, as its solutions.

Principal Solution

The solutions of a trigonometric equation that lie in the interval [0, 2π). For example, 2sin? = 1 , then the two values of sin? between 0 and 2π are  π/6 and 5π/6. Thus, π/6 and 5π/6 are the principal solutions of equation 2sin? = 1.

General Solution

As trigonometric functions are periodic, solutions are repeated within each period, so, trigonometric equations may have an infinite number of solutions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution.

Some Important General Solutions of Equations

\mathbf{Equations}

\mathbf{Solution}

\sin\theta=0

\theta=n \pi, \quad n \in \mathbb{I}

\cos\theta=0

\theta=(2 n+1) \frac{\pi}{2}, \quad n \in \mathbb{I}

\tan\theta=0

\theta=n \pi, \quad n \in \mathbb{I}

\sin\theta=1

\theta=(4 n+1) \frac{\pi}{2}, \quad n \in \mathbb{I}

\cos\theta=1

\theta=2 n \pi, \quad n \in \mathbb{I}

\sin\theta=-1

{\theta}=(4 n-1) \frac{\pi}{2}, \quad n \in \mathbb{I}

\cos\theta=-1

\theta=(2 n+1) \pi, \quad n \in \mathbb{I}

\cot\theta=0

\theta=(2 n+1) \frac{\pi}{2}, \quad n \in \mathbb{I}

2\sin^2 x+3\cos 2x-3=0\\ 2\sin^2 x+3(cos^2 x- \sin^2 x)-3=0\\ 3cos^2 x- \sin^2 x-3=0\\ 4cos^2 x- 4=0\\ \cos x=\pm 1\\ x=\{0, \pi, 2\pi \}

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