2q,b ii) \( \sin 20=0,64 \) vi) \( 3-3 \cos 20=2,28 \). 3. If \( 2 \tan \left(x+53^{\circ}\right)=-1,26 \), calculate solution
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To solve for \( x \) in the equation \( 2 \tan \left(x+53^{\circ}\right) = -1.26 \), first divide both sides by 2 to isolate the tangent: \[ \tan \left(x + 53^{\circ}\right) = -0.63 \] Next, to find \( x + 53^{\circ} \), take the arctangent: \[ x + 53^{\circ} = \tan^{-1}(-0.63) \] Calculating \( \tan^{-1}(-0.63) \), you'll find an angle in the fourth quadrant (approximately -32°). Since tangent is periodic and has a period of \( 180° \), we can find an additional solution by adding \( 180° \): \[ x + 53^{\circ} = -32° + 180° = 148° \] Now, solve for \( x \): 1. \( x + 53^{\circ} = -32° \) gives \( x = -32° - 53° = -85°\) 2. \( x + 53^{\circ} = 148° \) gives \( x = 148° - 53° = 95°\) Therefore, the two solutions for \( x \) are \( -85° \) and \( 95° \). Just remember when dealing with angles to keep track of which quadrant you're in when solving for \( x \)!