We are given the trigonometric equation:

\[
7\sin(x) + 2\cos(2x) = 4
\]

### Step 1: Express \(\cos(2x)\) in terms of \(\sin(x)\)
Using the double-angle identity for cosine:

\[
\cos(2x) = 1 - 2\sin^2(x)
\]

Now substitute this into the original equation:

\[
7\sin(x) + 2(1 - 2\sin^2(x)) = 4
\]

Simplify:

\[
7\sin(x) + 2 - 4\sin^2(x) = 4
\]

### Step 2: Rearrange the equation
Move all terms to one side of the equation:

\[
-4\sin^2(x) + 7\sin(x) + 2 = 4
\]

Subtract 4 from both sides:

\[
-4\sin^2(x) + 7\sin(x) - 2 = 0
\]

### Step 3: Solve the quadratic equation
We have a quadratic equation in terms of \(\sin(x)\):

\[
-4\sin^2(x) + 7\sin(x) - 2 = 0
\]

Let \(y = \sin(x)\), then:

\[
-4y^2 + 7y - 2 = 0
\]

Solve this quadratic using the quadratic formula:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = -4\), \(b = 7\), and \(c = -2\). Substituting these values:

\[
y = \frac{-7 \pm \sqrt{7^2 - 4(-4)(-2)}}{2(-4)}
\]
\[
y = \frac{-7 \pm \sqrt{49 - 32}}{-8}
\]
\[
y = \frac{-7 \pm \sqrt{17}}{-8}
\]

Thus, the two possible solutions for \(y\) (i.e., \(\sin(x)\)) are:

\[
y_1 = \frac{-7 + \sqrt{17}}{-8}, \quad y_2 = \frac{-7 - \sqrt{17}}{-8}
\]

### Step 4: Approximate the solutions
We can approximate the square root of 17:

\[
\sqrt{17} \approx 4.123
\]

Substitute this into the expressions for \(y\):

\[
y_1 = \frac{-7 + 4.123}{-8} \approx \frac{-2.877}{-8} \approx 0.3596
\]
\[
y_2 = \frac{-7 - 4.123}{-8} \

Discover how to solve the trigonometric equation 7sin(x) + 2cos(2x) = 4 step by step. This solution involves using the double-angle identity and the quadratic formula, suitable for students in Grades 10-12.

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