Solve the equation(2sin x-1)(2cos x+1)=0] within the interval [-180 <= x <= 180],[

Set Factors to Zero: First, we set each factor in the equation to zero because if the product of two factors is zero, at least one of the factors must be zero. So, we have 2sin x - 1 = 0 and 2cos x + 1 = 0.Solve First Equation: Now, let's solve the first equation: 2sin x - 1 = 0. Add 1 to both sides to isolate the term with sin x: 2sin x = 1.Find Angles for sin x: Next, divide both sides by 2 to solve for sin x: sin x = 1/2.Solve Second Equation: We look for angles where the sine is 1/2. These are x = 30 degrees and x = 150 degrees.Find Angles for cos x: Now, let's solve the second equation: 2cos x + 1 = 0. Subtract 1 from both sides to isolate the term with cos x: 2cos x = -1.Then, divide both sides by 2 to solve for cos x: cos x = -1/2.We look for angles where the cosine is -1/2. These are x = -120 degrees and x = 120 degrees.

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Solve the equation(2sin x-1)(2cos x+1)=0] within the interval [-180 <= x <= 180],[

Solve the equation $\newline$ $(2\sin x-1)(2\cos x+1)=0$ within the interval $-180 \leq x \leq 180$ ,

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Q. Solve the equation

\newline

(2\sin x-1)(2\cos x+1)=0

within the interval

-180 \leq x \leq 180

Set Factors to Zero: First, we set each factor in the equation to zero because if the product of two factors is zero, at least one of the factors must be zero. $\newline$ So, we have $2\sin x - 1 = 0$ and $2\cos x + 1 = 0$ .
Solve First Equation: Now, let's solve the first equation: $2\sin x - 1 = 0$ . Add $1$ to both sides to isolate the term with \ ext{sin} x: $2\sin x = 1$ .
Find Angles for $\sin x$ : Next, divide both sides by $2$ to solve for $\sin x$ : $\sin x = \frac{1}{2}$ .
Solve Second Equation: We look for angles where the sine is $\frac{1}{2}$ . These are $x = 30$ degrees and $x = 150$ degrees.
Find Angles for $\cos x$ : Now, let's solve the second equation: $2\cos x + 1 = 0$ . Subtract $1$ from both sides to isolate the term with $\cos x$ : $2\cos x = -1$ .
Find Angles for $\cos x$ : Now, let's solve the second equation: $2\cos x + 1 = 0$ . Subtract $1$ from both sides to isolate the term with $\cos x$ : $2\cos x = -1$ . Then, divide both sides by $2$ to solve for $\cos x$ : $\cos x = -\frac{1}{2}$ .
Find Angles for $\cos x$ : Now, let's solve the second equation: $2\cos x + 1 = 0$ . Subtract $1$ from both sides to isolate the term with $\cos x$ : $2\cos x = -1$ . Then, divide both sides by $2$ to solve for $\cos x$ : $\cos x = -\frac{1}{2}$ . We look for angles where the cosine is $-\frac{1}{2}$ . These are $x = -120$ degrees and $2\cos x + 1 = 0$ $0$ degrees.