Question 25

If $$3 \cos^2 x - 2 \sin^2 x = -0.75$$ and $$0° \leq x \leq 90°$$, then $$x$$ =

Solution

$$3 \cos^2 x - 2 \sin^2 x = -0.75$$

$$3 \cos^2 x - 2 (1-\cos^2 x) = -0.75$$

$$5 \cos^2 x - 2 = -0.75$$

$$5 \cos^2 x = 1.25$$

$$ \cos^2 x = \frac{1.25}{5}$$

$$ \cos^2 x = 0.25 $$

$$ \cos x = 0.5 $$

$$ \cos x = \cos60 ^\circ $$ $$[\cos60^\circ = 0.5, 0° \leq x \leq 90°]$$

$$ x =60 ^\circ $$

Option B is correct.

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