Step | Comment |
---|---|
Step 1 | Identify the equation, which is \(2cos^2(x) + sin(x) – 1 = 0\). |
Step 2 | Use the Pythagorean identity \(sin^2(x) + cos^2(x) = 1\) to express \(cos^2(x)\) as \(1 – sin^2(x)\). |
Step 3 | Substitute \(1 – sin^2(x)\) for \(cos^2(x)\) in the equation to get \(2(1 – sin^2(x)) + sin(x) – 1 = 0\). |
Step 4 | Simplify the equation to get \(2 – 2sin^2(x) + sin(x) – 1 = 0\). |
Step 5 | Further simplify the equation to get \(2sin^2(x) – sin(x) – 1 = 0\). |
Step 6 | This is a quadratic equation in terms of \(sin(x)\). We can solve it by factoring: \(2sin^2(x) – sin(x) – 1 = 0\) factors to \((2sin(x) + 1)(sin(x) – 1) = 0\). |
Step 7 | Set each factor equal to zero and solve for \(sin(x)\) to get \(sin(x) = -1/2\) or \(sin(x) = 1\). |
Step 8 | Find the values of \(x\) that satisfy \(sin(x) = -1/2\). These are \(x = 7π/6, 11π/6\). |
Step 9 | Find the values of \(x\) that satisfy \(sin(x) = 1\). This is \(x = π/2\). |
Step 10 | Note that these are the solutions in the first period \([0, 2π)\). The general solutions can be found by adding \(2πn\) to each of these solutions, where \(n\) is an integer. |
Step | Comment |
---|---|
Step 1 | Identify the equation, which is \(3cos^2(x) – sin(x) – 2 = 0\). |
Step 2 | Use the Pythagorean identity \(sin^2(x) + cos^2(x) = 1\) to express \(cos^2(x)\) as \(1 – sin^2(x)\). |
Step 3 | Substitute \(1 – sin^2(x)\) for \(cos^2(x)\) in the equation to get \(3(1 – sin^2(x)) – sin(x) – 2 = 0\). |
Step 4 | Simplify the equation to get \(3 – 3sin^2(x) – sin(x) – 2 = 0\). |
Step 5 | Further simplify the equation to get \(3sin^2(x) + sin(x) – 1 = 0\). |
Step 6 | This is a quadratic equation in terms of \(sin(x)\). We can solve it by factoring: \(3sin^2(x) + sin(x) – 1 = 0\) factors to \((3sin(x) – 1)(sin(x) + 1) = 0\). |
Step 7 | Set each factor equal to zero and solve for \(sin(x)\) to get \(sin(x) = 1/3\) or \(sin(x) = -1\). |
Step 8 | Find the values of \(x\) that satisfy \(sin(x) = 1/3\). This is \(x = arcsin(1/3)\). |
Step 9 | Find the values of \(x\) that satisfy \(sin(x) = -1\). This is \(x = 3π/2\). |
Step 10 | Note that these are the solutions in the first period \([0, 2π)\). The general solutions can be found by adding \(2πn\) to each of these solutions, where \(n\) is an integer. |