BenoguigoliB

2021-08-07

Solving Basic Trigonometric Equations by Factoring Solve the given equation.

$3{\mathrm{sin}}^{2}\theta -7\mathrm{sin}\theta +2=0$

Macsen Nixon

Skilled2021-08-08Added 117 answers

Approach:

The domain of the trigonometry function of$\mathrm{sin}\theta$ is lies between [-1,1]. No solution exists beyond this domain. Sine has period $2\pi$ , we find solution in any interval of length $2\pi$ . SIne function is positive in first and second quadrant.

Calculation:

Consider the trigonometry equation.

$3{\mathrm{sin}}^{2}\theta -7\mathrm{sin}\theta +2=0$

The factor of the given equation is obtained by,

$3{\mathrm{sin}}^{2}\theta -7\mathrm{sin}\theta +2=0$

$(\mathrm{sin}\theta -2)(3\mathrm{sin}\theta -1)=0$

The factors of above equation are,

$\mathrm{sin}\theta -2=0\dots \dots \left(1\right)$

$3\mathrm{sin}\theta -1=0\dots \dots \left(2\right)$

The solution obtained for the factor in which sine function involved so we will get the solution in the interval of$[0,2\pi ]$ .

Consider equation (1).

$\mathrm{sin}\theta -2=0$

The above equation has no solution because value lies outside the domain.

Add 1 both sides in equation (2).

$3\mathrm{sin}\theta -1=0$

$\theta ={\mathrm{sin}}^{1}\left(\frac{1}{3}\right)$

$\theta =0.34,2.80$

Here, the angles are in radian.

The sine has period,$2\pi$ So we get all solutions of the equation by adding integer multiples of $2\pi$ to these solutions:

S$\theta =0.34+2k\pi$

$\theta =2.80+2k\pi$

Therefore, the solutions of the trigonometry equation$3{\mathrm{sin}}^{2}\theta -7\mathrm{sin}\theta +2=0$ are $\theta =0.34+2k\pi$ and $\theta =2.80+2k\pi$ .

Concusion:

Thus, the solutions of the trigonometry equation$3{\mathrm{sin}}^{2}\theta -7\mathrm{sin}\theta +2=0$ are $\theta =0.34+2k\pi$ and $\theta =2.80+2k\pi$ .

The domain of the trigonometry function of

Calculation:

Consider the trigonometry equation.

The factor of the given equation is obtained by,

The factors of above equation are,

The solution obtained for the factor in which sine function involved so we will get the solution in the interval of

Consider equation (1).

The above equation has no solution because value lies outside the domain.

Add 1 both sides in equation (2).

Here, the angles are in radian.

The sine has period,

S

Therefore, the solutions of the trigonometry equation

Concusion:

Thus, the solutions of the trigonometry equation

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