Solving a Trig Equation Using an Identity

When an equation mixes trig functions, the strategy is to reduce it to one function. Here $\tan x = \frac{\sin x}{\cos x}$ and $\cos^2 x = 1 - \sin^2 x$ turn everything into a quadratic in $\sin x$.

Solve $3\tan x = 2\cos x$ for $0^\circ \le x \le 360^\circ$. [6]

The working

Step 1, write $\tan x$ as $\frac{\sin x}{\cos x}$ and clear the fraction (multiply by $\cos x$, valid since $\cos x \ne 0$ at a solution): $3\sin x = 2\cos^2 x \quad \text{(M1)}$

Step 2, replace $\cos^2 x$ with $1 - \sin^2 x$: $3\sin x = 2(1 - \sin^2 x) \implies 2\sin^2 x + 3\sin x - 2 = 0 \quad \text{(M1, A1)}$

Step 3, factorise the quadratic in $\sin x$: $(2\sin x - 1)(\sin x + 2) = 0 \quad \text{(M1)}$ $\sin x = \tfrac{1}{2} \quad \text{or} \quad \sin x = -2 \ (\text{rejected, since } |\sin x| \le 1)$

Step 4, solve $\sin x = \frac{1}{2}$ over the range: $x = 30^\circ, \ 150^\circ \quad \text{(A1, A1)}$

Where the marks are won and lost

• Reject $\sin x = -2$ explicitly. Stating why (sine is between $-1$ and $1$) shows the examiner you understand the constraint.
• Find both angles. $\sin x = \tfrac12$ is positive in the first and second quadrants, giving $30^\circ$ and $180^\circ - 30^\circ = 150^\circ$.
• Keep the quadratic tidy: bring everything to one side so it reads $2\sin^2 x + 3\sin x - 2 = 0$ before factorising.

Common mistakes

• Only giving $x = 30^\circ$ and missing $150^\circ$.
• Using $\cos^2 x = \sin^2 x - 1$ (wrong sign) instead of $1 - \sin^2 x$.
• Cancelling $\cos x$ carelessly and losing solutions or introducing errors.

Topic home: Trigonometry pillar. More: Worked examples.