how can i find all the exact solutions to the equation : 2cos^2x + 3sinx = 3
the solutions have to be between [0,2pi)
assuming that 2cos^2x means 2(cos(x))^2
substitute using
(cos(x))^2 = 1 - (sin(x))^2
2(1 - (sin(x))^2) + 3sin(x) = 3
2 - 2(sin(x))^2 + 3sin(x) = 3
for clarity, let u = sin(x)
2 - 2u^2 + 3u = 3
rearranging...
2u^2 - 3 u + 1 = 0
Solve this quadratic to find the values of sin x that are solutions. Finally, determine the angles in the range that have that value of sin(x).
To find the exact solutions to the equation 2cos^2x + 3sinx = 3, we can follow these steps:
Step 1: Express everything in terms of sinx.
Since cos^2x = 1 - sin^2x, we can rewrite the equation as 2(1 - sin^2x) + 3sinx = 3.
Simplifying further, we get:
2 - 2sin^2x + 3sinx = 3
-2sin^2x + 3sinx - 1 = 0
Step 2: Rearrange the equation.
Rearrange the equation to bring all terms to one side:
-2sin^2x + 3sinx - 1 = 0
Step 3: Factor the quadratic equation.
To factor the equation, we need to find two numbers that multiply together to give -2 and add up to 3. The numbers -1 and 2 fit this condition:
(-2sinx + 1)(sinx - 1) = 0
Set each factor equal to zero:
-2sinx + 1 = 0 or sinx - 1 = 0
Step 4: Solve each factor separately.
Solving -2sinx + 1 = 0:
-2sinx = -1
sinx = 1/2
Solving sinx - 1 = 0:
sinx = 1
Step 5: Find the solutions within the specified interval [0, 2pi).
sinx = 1/2 has two solutions within the interval [0, 2pi): pi/6 and 5pi/6.
sinx = 1 also has a solution within the interval [0, 2pi): pi/2.
Therefore, the exact solutions to the equation 2cos^2x + 3sinx = 3 within the interval [0, 2pi) are:
x = pi/6, 5pi/6, and pi/2.