Score: $1 / 2$
Penalty: none
Question
Solve the trigonometric equation for all values $0 \leq x < 2 π$ .
$4 \sin^{2} x - 1 = 0$
Answer Attempt 1 out of 2
Additional Solution
No Solution
$x =$

Answer

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Answer

Final Answer: The solutions to the equation $4 \sin^{2} x - 1 = 0$ in the interval $0 \leq x < 2 π$ are $x = 0.523598775598299, 2.61799387799149, 3.66519142918809$ .

Steps

Step 1 ：The given equation is a quadratic equation in terms of $\sin^{2} x$ . We can solve it by setting it equal to zero and finding the roots.

Step 2 ：The roots will give us the values of $\sin x$ , and we can then find the corresponding values of $x$ in the interval $0 \leq x < 2 π$ .

Step 3 ：The solutions to the equation are $x = 0.523598775598299, 2.61799387799149, 3.66519142918809$ . These are the values of $x$ in the interval $0 \leq x < 2 π$ that satisfy the equation $4 \sin^{2} x - 1 = 0$ .

Step 4 ：Final Answer: The solutions to the equation $4 \sin^{2} x - 1 = 0$ in the interval $0 \leq x < 2 π$ are $x = 0.523598775598299, 2.61799387799149, 3.66519142918809$ .

Score: 1/2Penalty: noneQuestionSolve the trigonometric equation for all values 0≤x<2π.4sin2⁡x−1=0Answer Attempt 1 out of 2Additional SolutionNo Solutionx=

Answer

Popular Problems

Score: $1 / 2$
Penalty: none
Question
Solve the trigonometric equation for all values $0 \leq x < 2 π$ .
$4 \sin^{2} x - 1 = 0$
Answer Attempt 1 out of 2
Additional Solution
No Solution
$x =$