Solve, $4+\sqrt{3}-2(1+\sqrt{3}) \sin x=4 \cos ^2 x$

To solve the equation 4 + 3 ​ − 2 ( 1 + 3 ​ ) sin x = 4 cos 2 x , we substitute cos 2 x with 1 − sin 2 x to form a quadratic equation in terms of sin x . Using the quadratic formula, we find possible values for sin x and solve for x to obtain solutions in the form of x = 6 π ​ + 2 nπ , 3 π ​ + 2 nπ , 3 2 π ​ + 2 nπ , 6 5 π ​ + 2 nπ .
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Answered by Anonymous | 2025-07-04

Rewrite the equation using the identity cos 2 x = 1 − sin 2 x and simplify to get a quadratic equation in terms of sin x .
Let y = sin x and solve the quadratic equation 4 y 2 − 2 ( 1 + 3 ​ ) y + 3 ​ = 0 for y using the quadratic formula.
Find the two possible values for y : y 1 ​ = 2 3 ​ ​ and y 2 ​ = 2 1 ​ .
Solve for x in each case: x = 6 π ​ + 2 nπ , x = 3 π ​ + 2 nπ , x = 3 2 π ​ + 2 nπ , and x = 6 5 π ​ + 2 nπ , where n is an integer. x = 6 π ​ + 2 nπ , 3 π ​ + 2 nπ , 3 2 π ​ + 2 nπ , 6 5 π ​ + 2 nπ ​

Explanation

Problem Setup We are given the equation 4 + 3 ​ − 2 ( 1 + 3 ​ ) sin x = 4 cos 2 x . Our goal is to solve for x .

Using Trigonometric Identity First, we rewrite the equation using the identity cos 2 x = 1 − sin 2 x . This gives us


4 + 3 ​ − 2 ( 1 + 3 ​ ) sin x = 4 ( 1 − sin 2 x ) .

Simplifying to Quadratic Form Expanding and simplifying the equation, we get

4 + 3 ​ − 2 sin x − 2 3 ​ sin x = 4 − 4 sin 2 x .
Rearranging the terms, we obtain a quadratic equation in terms of sin x :
4 sin 2 x − 2 ( 1 + 3 ​ ) sin x + 3 ​ = 0 .

Applying Quadratic Formula Let y = sin x . The equation becomes 4 y 2 − 2 ( 1 + 3 ​ ) y + 3 ​ = 0 . We can solve this quadratic equation for y using the quadratic formula:

y = 2 a − b ± b 2 − 4 a c ​ ​ , where a = 4 , b = − 2 ( 1 + 3 ​ ) , and c = 3 ​ .

Calculating the Discriminant The discriminant is D = b 2 − 4 a c = [ − 2 ( 1 + 3 ​ ) ] 2 − 4 ( 4 ) ( 3 ​ ) = 4 ( 1 + 2 3 ​ + 3 ) − 16 3 ​ = 4 ( 4 + 2 3 ​ ) − 16 3 ​ = 16 + 8 3 ​ − 16 3 ​ = 16 − 8 3 ​ = 8 ( 2 − 3 ​ ) .

Finding Solutions for y Now we find the solutions for y :


y = 8 2 ( 1 + 3 ​ ) ± 8 ( 2 − 3 ​ ) ​ ​ = 8 2 ( 1 + 3 ​ ) ± 2 2 ( 2 − 3 ​ ) ​ ​ = 4 1 + 3 ​ ± 4 − 2 3 ​ ​ ​ .
Notice that 4 − 2 3 ​ = ( 3 ​ − 1 ) 2 , so 4 − 2 3 ​ ​ = ( 3 ​ − 1 ) 2 ​ = ∣ 3 ​ − 1∣ = 3 ​ − 1 .
Thus, y = 4 1 + 3 ​ ± ( 3 ​ − 1 ) ​ .

Two Possible Values for y The two possible values for y are:

y 1 ​ = 4 1 + 3 ​ + 3 ​ − 1 ​ = 4 2 3 ​ ​ = 2 3 ​ ​ and y 2 ​ = 4 1 + 3 ​ − 3 ​ + 1 ​ = 4 2 ​ = 2 1 ​ .

Finding sin x Since y = sin x , we have sin x = 2 3 ​ ​ or sin x = 2 1 ​ .

Solving for x Now we solve for x in each case:


For sin x = 2 3 ​ ​ , x = 3 π ​ + 2 nπ or x = 3 2 π ​ + 2 nπ , where n is an integer.
For sin x = 2 1 ​ , x = 6 π ​ + 2 nπ or x = 6 5 π ​ + 2 nπ , where n is an integer.

Final Solutions Therefore, the solutions for x are x = 6 π ​ + 2 nπ , x = 3 π ​ + 2 nπ , x = 3 2 π ​ + 2 nπ , and x = 6 5 π ​ + 2 nπ , where n is an integer.

Examples
Trigonometric equations are used in physics to model oscillations and wave phenomena, such as the motion of a pendulum or the propagation of light waves. Solving these equations allows us to predict the behavior of these systems over time. For example, in electrical engineering, trigonometric functions describe alternating current (AC) circuits, and solving trigonometric equations helps determine the current and voltage at different points in the circuit. In navigation, these equations are used to calculate angles and distances, essential for determining a ship's or aircraft's position.

Answered by GinnyAnswer | 2025-07-04