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Find all values of x in the interval [0, 2π] that satisfy the equation. 7 sin(2x) = 7 cos(x)

User Mistertim
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1 Answer

5 votes

Answer:

x = {π/6, π/2, 5π/6, 3π/2}

Explanation:

The equation can be solved using a double-angle trig identity and factoring.

Simplify

Dividing the equation by 7 and substituting for sin(2x), we have ...

7sin(2x) = 7cos(x)

sin(2x) = cos(x)

2sin(x)cos(x) = cos(x)

2sin(x)cos(x) -cos(x) = 0

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

Zero product rule

The product of factors is zero when one or more of the factors is zero.

cos(x) = 0 ⇒ x = {π/2, 3π/2}

2sin(x) -1 = 0 ⇒ x = arcsin(1/2) = {π/6, 5π/6}

Solutions in the given interval are ...

x = {π/6, π/2, 5π/6, 3π/2}

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Additional comment

When the equation is of the form f(x) = 0, then the x-intercepts of f(x) are its solutions. We can rearrange this one to ...

sin(2x) -cos(x) = 0

The solutions identified above match those shown in the graph.

Find all values of x in the interval [0, 2π] that satisfy the equation. 7 sin(2x) = 7 cos-example-1
User Flea
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4.0k points