To solve: The equation 4 \cos \theta \sin \theta + 3 \cos \theta = 0

BolkowN

BolkowN

Answered question

2021-07-31

To solve:
The equation 4cosθsinθ+3cosθ=0

Answer & Explanation

Willie

Willie

Skilled2021-08-01Added 95 answers

Approach:
The domain of the trigonometry function of cosθ is lies between [1,1]. No solution exists beyond this domain.
Obtain factor of given equation and simply the factor equations to obtain solutions.
Cosine and sine have period 2π, thus find the solution in any interval of length 2π.
Sine function is positive in first and second quadrant. Cosine function is positive in first and fourth quadrant.
Calculation:
Consider the trigonometry equation.
4cosθsinθ+3cosθ=0
The factors of above equation are obtained by,
4cosθsinθ+3cosθ=0
cosθ(4sinθ+3)=0
The factors are,
4sinθ+3=0(1)
cosθ=0(2)
The solution obtained for the factor in which sine and cosine functions are involved so we will get the solution in the interval of [0,2π]
Substract 3 both sides in equation (1).
4sinθ=3
Divide by 4 both sides in equation (1).
sinθ=34
Multiple by sin1 both sides in equation (1).
sin1sinθ=sin1(34)
θ=sin1(34)
=4,5.435
Here solutions are in radian.
The solution repeats value of the equation at every length of π in the interval [0,2π].
We will get all solutions of the equation by adding integer multiples of π to these solutions:
θ=4+2kπ
θ=5.435+2kπ
Consider the factor in equation (2).
cosθ=0
θ=π2,3π2
The solution repeats value of the equation at every length of π in the interval [0,2π]
We will get all solutions of the equation by adding integer multiples of π to these solutions: θ=π2+kπ
Therefore, the solutions of the trigonometry equation 4cosθsinθ+3cosθ=0 are θ=π2+kπ,θ=4+2kπ and θ=5.435+2kπ
Conclusion:
Hence, the solutions of the trigonometry equation 4cosθsinθ+3cosθ=0 are θ=π2+kπ,θ=4+2kπ and θ=5.435+2kπ

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