tan(x)=5 sin(x) for interval -π < x < π
tan(x)=sin(x)/cos(x)
sin(x)/cos(x)=5sin(x) Divide withsin(x)
1/cos(x)=5
1=5*cos(x) Divide with 5
cos(x)=1/5
cos(x)=0.2
OR
1/cos(x)=5
sec(x)=5
The only choices were A) 0, 1.571 B) -1.571, 0, 1.571 C) -1.369, 0, 1.369 D) 0, 1.369
tan(0)=0
sin(0)=0
tan(0)=5*sin(0)
0=0
arccos(0.2)=1.3694384 radians
tan(-alpha)= -tan(alpha)
sin(-alpha)= -sin(alpha)
tan(-1.3694384)=5 sin(-1.3694384)
Answer C) C) -1.369,0,1.369 is correct
To solve the equation tan(x) = 5 sin(x) for the given interval -π < x < π, follow these steps:
Step 1: Rewrite the equation using the identity tan(x) = sin(x) / cos(x).
sin(x) / cos(x) = 5 sin(x)
Step 2: Multiply both sides of the equation by cos(x) to eliminate the denominator.
sin(x) = 5 sin(x) cos(x)
Step 3: Rearrange the equation by moving all terms to one side.
0 = 5 sin(x) cos(x) - sin(x)
Step 4: Factor out sin(x) from both terms.
0 = sin(x) (5 cos(x) - 1)
Step 5: Set each factor equal to zero and solve for x.
sin(x) = 0 or 5 cos(x) - 1 = 0
For sin(x) = 0, the solutions are x = 0, π, -π. These values of x satisfy the equation sin(x) = 0 in the given interval.
For 5 cos(x) - 1 = 0, add 1 to both sides and divide by 5 to isolate cos(x).
cos(x) = 1/5
To find the values of x that satisfy this equation, use the inverse cosine function:
x = cos^(-1)(1/5) ≈ 1.3694 and x = -cos^(-1)(1/5) ≈ -1.3694
However, we need to check whether these solutions are within the given interval -π < x < π.
Since 1.3694 is greater than π, it is not within the given interval. However, -1.3694 is within the interval.
Therefore, the solutions for the equation tan(x) = 5 sin(x) in the interval -π < x < π are x = 0, π, -π, and approximately x = -1.3694.