Keep only valid sine values
Solve $2\cos^2x-3\sin x=0$ for $0\le x<2\pi$.
$\sin x=-2$ is impossible. From $\sin x=\frac12$, the interval gives
Select useful identities, solve on a stated interval, and retain every valid trigonometric solution.
Why this matters. The main difficulty is usually not algebra alone. It is choosing a form that exposes the equation without losing or inventing solutions.
Choose the version that reduces the number of different functions or angles in the equation.
Checkpoint
Correct. It turns the equation into a quadratic in $\sin x$.
Not quite. Aim for one trigonometric function. The other identities retain mixed functions or angles.
It turns the equation into a quadratic in $\sin x$.
Example 1
Keep only valid sine values
Solve $2\cos^2x-3\sin x=0$ for $0\le x<2\pi$.
$\sin x=-2$ is impossible. From $\sin x=\frac12$, the interval gives
Dividing by $\sin x$ or $\cos x$ can discard cases where that factor is zero. Factor first and solve each branch.
Checkpoint
Correct. A product is zero when either factor is zero.
Not quite. Dividing by $\sin x$ would lose the entire $\sin x=0$ branch.
A product is zero when either factor is zero.
Example 2
Exact evaluation
Find $\sin75^\circ$ exactly.
The signs follow directly from the addition identity.
Checkpoint
Correct. $0\le2x<2\pi$, so $2x=\pi/2,3\pi/2$.
Not quite. Transform the interval as well as the equation when the angle is multiplied.
$0\le2x<2\pi$, so $2x=\pi/2,3\pi/2$.
Example 3
Solve $\sin x+\cos x=1$
Since $\sin x+\cos x=\sqrt2\sin(x+\pi/4)$, solve
For $0\le x<2\pi$, this gives $x+\pi/4=\pi/4,3\pi/4$ modulo $2\pi$, hence
Substitution confirms both values.
Five questions to check the main decisions from this lesson.
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Rearrange the Pythagorean identity.
Rearrange the Pythagorean identity.
This is the double-angle identity for sine.
This is the double-angle identity for sine.
Division assumes the divisor is non-zero.
Division assumes the divisor is non-zero.
$2\pi$ is excluded by the half-open interval.
$2\pi$ is excluded by the half-open interval.
$2x=2k\pi$, so $x=k\pi$.
$2x=2k\pi$, so $x=k\pi$.
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